Question 512 Marks
If $a, b, c$ are all nonzero and $a + b + c = 0$, prove that: $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}=3$
Answer$a+b+c=0 \Rightarrow a^3+b^3+c^3=3 a b c$ Thus, We have: $\Big(\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}\Big)=\frac{\text{a}^3+\text{b}^3+\text{c}^3}{\text{abc}}$ $=\frac{3\text{abc}}{\text{abc}}$ $=3$
View full question & answer→Question 522 Marks
Factorise: $2\text{x}^2+3\sqrt{3}\text{x}+3$
Answer$2\text{x}^2+3\sqrt{3}\text{x}+3$ $2\text{x}^2+2\sqrt{3}\text{x}+\sqrt{3}\text{x}+3$ $=2\text{x}\big(\text{x}+\sqrt{3}\big)+\sqrt{3}\big(\text{x}+\sqrt{3}\big)$ $=\big(\text{x}+\sqrt{3}\big)\big(2\text{x}+\sqrt{3}\big)$
View full question & answer→Question 532 Marks
Expand: $\Big(3\text{x}-\frac{5}{\text{x}}\Big)^3$
Answer $\Big(3\text{x}-\frac{5}{\text{x}}\Big)^3$
$=(3\text{x})^3-\Big(\frac{5}{\text{x}}\Big)^3-3(3\text{x})^2\Big(\frac{5}{\text{x}}\Big)+3(3\text{x})\Big(\frac{5}{\text{x}}\Big)^2$
$=27\text{x}^3-\frac{125}{\text{x}^3}-135\text{x}+\frac{225}{\text{x}}$
View full question & answer→Question 542 Marks
Evaluate:
$(107)^2$
Answer$(107)^2=(100+7)^2$
$=(100)^2+2 \times(100) \times(7)+(7)^2$
$=10000+1400+49$
$=11449$
View full question & answer→Question 552 Marks
Expand:
$(5 a-3 b)^3$
Answer$(5 a-3 b)^3$
$=(5 a)^3-(3 b)^3-3(5 a)^2(3 b)+3(5 a)(3 b)^2$
$=125 a^3-27 b^3-225 a^2 b+135 a b^2$
View full question & answer→Question 562 Marks
Factorise:
$x^2-32 x-105$
Answer$x^2-32 x-105$
$=x^2-35 x+3 x-105$
$=x(x-35)+3(x-35)$
$=x(x-35)(x+3)$
View full question & answer→Question 572 Marks
Factorise:
$6\sqrt{3}\text{x}^2-47\text{x}+5\sqrt{3}$
Answer $6\sqrt{3}\text{x}^2-47\text{x}+5\sqrt{3}$
$=6\sqrt{3}\text{x}^2-45\text{x}-2\text{x}+5\sqrt{3}$
$=3\sqrt{3}\text{x}\big(2\text{x}-5\sqrt{3}\big)-1\big(2\text{x}-5\sqrt{3}\big)$
$=\big(2\text{x}-5\sqrt{3}\big)\big(3\sqrt{3}\text{x}-1\big)$
View full question & answer→Question 582 Marks
Factorise:
$(a+b)^3-a-b$
Answer$(a+b)^3-a-b$
$=(a+b)^3-(a+b)$
$=(a+b)\left[(a+b)^2-1^2\right]\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
$=(a+b)(a+b+1)(a+b-1)$
View full question & answer→Question 592 Marks
Factorise: $2\sqrt{2}\text{a}^3+16\sqrt{2}\text{b}^3+\text{c}^3-12\text{abc}$
Answer$x =2\sqrt{2}\text{a}^3+16\sqrt{2}\text{b}^3+\text{c}^3-12\text{abc}$ $=\big(\sqrt{2}\text{a}\big)^3+\big(2\sqrt{2}\text{b}\big)^3+\text{c}^3-3\times\big(\sqrt{2}\text{a}\big)\times\big(2\sqrt{2}\text{b}\big)\times(\text{c})$ $=\big(\sqrt{2}\text{a}+2\sqrt{2}\text{b}+\text{c}\big)\Big[\big(\sqrt{2}\text{a}\big)^2+(2\sqrt{2}\text{b})^2+\text{c}^2\\-\big(\sqrt{2}\text{a}\big)\times\big(2\sqrt{2}\text{b}\big)-\big(2\sqrt{2}\text{b}\big)\times(\text{c})-\big(\sqrt{2}\text{a}\big)\times(\text{c})\Big]$ $=\big(\sqrt{2}\text{a}+2\sqrt{2}\text{b}+\text{c}\big)\big(2\text{a}^2+8\text{b}^2+\text{c}^2-4\text{ab}-2\sqrt{2\text{bc}}-\sqrt{2}\text{ac}\big)$
View full question & answer→Question 602 Marks
Factorise: $\text{x}^2+\sqrt{2}\text{x}-24$
Answer$\text{x}^2+\sqrt{2}\text{x}-24$ $=\text{x}^2+4\sqrt{2}\text{x}-3\sqrt{2}\text{x}-24$ $=\text{x}(\text{x}+4\sqrt{2})-3\sqrt{2}(\text{x}+4\sqrt{2})$ $=(\text{x}-4\sqrt{2})(\text{x}-3\sqrt{2})$
View full question & answer→Question 612 Marks
Factorise:
$10 x^2-9 x-7$
Answer$10 x^2-9 x-7$
$=10 x^2+5 x-14 x-7$
$=5 x(2 x+1)-7(2 x+1)$
$=(2 x+1)(5 x-7)$
View full question & answer→Question 622 Marks
Factorise:
$x^2-21 x+90$
Answer$x^2-21 x+90$
$=x^2-6 x-15 x+90$
$=x(x-6)-15(x-6)$
$=(x-6)(x-15)$
View full question & answer→Question 632 Marks
Factorise:
$1029-3 x^3$
Answer$1029-3 x^3$
$=3\left(343-x^3\right)$
$=3\left[(7)^3-x^3\right]$
$=3\left[(7-x)\left(7^2+7 x+x^2\right)\right]$
$=3(7-x)\left(49+7 x+x^2\right)$
View full question & answer→Question 642 Marks
Factorise:
$x^2+11 x+30$
Answer$x^2+11 x+30$
$=x^2+6 x+5 x+30$
$=x(x+6)+5(x+6)$
$=(x+6)(x+5)$
View full question & answer→Question 652 Marks
Factorise:
$25 x^2-10 x+1-36 y^2$
Answer$25 x^2-10 x+1-36 y^2$
$=\left(25 x^2-10 x+1\right)-36 y^2$
$=\left[(5 x)^2-2(5 x)(1)+(1)^2\right]-(6 y)^2$
$=(5 x-1)^2-(6 y)^2$
$=(5 x-1-6 y)(5 x-1+6 y)$
View full question & answer→Question 662 Marks
Factorise: $\frac{3}{2}\text{x}^2+16\text{x}+10$
Answer$\frac{3}{2}\text{x}^2+16\text{x}+10$ $=\frac{3}{2}\text{x}^2+\text{x}+15\text{x}+10$ $=\frac{\text{x}}{2}(3\text{x}+2)+5(3\text{x}+2)$ $(3\text{x}+2)\Big(\frac{\text{x}}{2}+5\Big)$
View full question & answer→Question 672 Marks
Factorise:
$x-8 x y^3$
Answer$x-8 x y^3$
$=x\left(1-8 y^3\right)$
$=x\left[(1)^3-(2 y)^3\right]$
$=x(1-2 y)\left[(1)^2+1 \times 2 y+(2 y)^2\right] \text { Since } a^3-b^3=(a-b)\left(a^2+a \times b+b^2\right)$
$=x(1-2 y)\left(1+2 y+4 y^2\right)$
View full question & answer→Question 682 Marks
Factorise: $\text{x}^4+\frac{4}{\text{x}^4}$
Answer$\text{x}^4+\frac{4}{\text{x}^4}$ $=\text{x}^4+\frac{4}{\text{x}^4}+4-4$ $=\big(\text{x}^2\big)^2+\Big(\frac{2}{\text{x}^2}\Big)^2+2\big(\text{x}^2\big)\Big(\frac{2}{\text{x}^2}\Big)-2^2$ $=\bigg[\big(\text{x}^2\big)^2+\Big(\frac{2}{\text{x}^2}\Big)^2+2\big(\text{x}^2\big)\Big(\frac{2}{\text{x}^2}\Big)\bigg]-2^2$ $=\Big[\text{x}^2+\frac{2}{\text{x}^2}\Big]^2-2^2$ $=\Big(\text{x}^2+\frac{2}{\text{x}^2}+2\Big)\Big(\text{x}^2+\frac{2}{\text{x}^2}-2\Big)$
View full question & answer→Question 692 Marks
Factorise:
$x^6-729$
Answer$x^6-729$
$=\left(x^2\right)^3-(9)^3$
$=\left(x^2-9\right)\left[\left(x^2\right)^2+x^2 \times 9+(9)^2\right] \text { Since } a^3-b^3=(a-b)\left(a^2+a \times b+b^2\right)$
$=\left(x^2-9\right)\left(x^4+9 x^2+81\right)$
$=(x+3)(x-3)\left[\left(x^2+9\right)^2-(3 x)^2\right]$
$=(x+3)(x-3)\left(x^2+3 x+9\right)\left(x^2-3 x+9\right)$
View full question & answer→Question 702 Marks
Factorise:
$27 a^3-b^3+8 c^3-18 a b c$
Answer$27 a^3-b^3+8 c^3-18 a b c$
$=(3 a)^3+(-b)^3+(2 c)^3-3 \times(3 a) \times(-b) \times(2 c)$
$=[3 a+(-b)+2 c]\left[(3 a)^2+(-b)^2+(2 c)^2-3 a(-b) 2 c-3 a \times 2 c\right]$
$=(3 a-b+2 c)\left(9 a^2+b^2+4 c^2+3 a b+2 b c-6 a c\right)$
View full question & answer→Question 712 Marks
Factorise: $1+\frac{27}{125}\text{a}^3+\frac{9\text{a}}{5}+\frac{27\text{a}^2}{25}$
Answer$1+\frac{27}{125}\text{a}^3+\frac{9\text{a}}{5}+\frac{27\text{a}^2}{25}$ $=(1)^3+\Big(\frac{3}{5}\text{a}\Big)^3+3(1)^2\Big(\frac{3}{5}\text{a}\Big)+3(1)\Big(\frac{3}{5}\text{a}\Big)^2$ $=\Big(1+\frac{3}{5}\text{a}\Big)^3$ Hence, factorisation of $1+\frac{27}{125}\text{a}^3+\frac{9\text{a}}{5}+\frac{27\text{a}^2}{25}$ is $=\Big(1+\frac{3}{5}\text{a}\Big)^3$
View full question & answer→Question 722 Marks
Factorise: $21\text{x}^2-2\text{x}+\frac{1}{21}$
Answer$21\text{x}^2-2\text{x}+\frac{1}{21}$ $21\text{x}^2-\text{x}-\text{x}+\frac{1}{21}$ $=21\text{x}\Big(\text{x}-\frac{1}{21}\Big)-1\Big(\text{x}-\frac{1}{21}\Big)$ $=\Big(\text{x}-\frac{1}{21}\Big)(21\text{x}-1)$
View full question & answer→Question 732 Marks
Factorise:
$a^3+0.008$
Answer$a^3+0.008$
$=(a)^3+(0.2)^3$
$=(a+0.2)\left[(a)^2-a \times 0.2+(0.2)^2\right] \text { Since } a^3+b^3=(a+b)\left(a^2-a \times b+b^2\right)$
$=(a+0.2)\left(a^2-0.2 a+0.04\right)$
View full question & answer→Question 742 Marks
Expand:
$(2 b-b+c)^2$
Answer$(2 b-b+c)^2=[(2 a)+(-b)+(c)]^2$
$=(2 a)^2+(-b)^2+(c)^2+2(2 a)(-b)+2(-b)(c)+4(a)(c)$
$=14 a^2+b^2+c^2-4 a b-2 b c+4 a c$
View full question & answer→Question 752 Marks
Factorise:
$\sqrt{3}\text{x}^2-10\text{x}+8\sqrt{3}$
Answer $\sqrt{3}\text{x}^2-10\text{x}+8\sqrt{3}$
$=\sqrt{3}\text{x}^2+4\text{x}+6\text{x}+8\sqrt{3}$
$=\text{x}\big(\sqrt{3}\text{x}+4\big)+2\sqrt{3}\big(\sqrt{3}\text{x}+4\big)$
$=\big(\sqrt{3}\text{x}+4\big)\big(\text{x}+2\sqrt{3}\big)$
View full question & answer→Question 762 Marks
Factorise: $\text{x}^2+5\sqrt{5}\text{x}+30$
Answer$\text{x}^2+5\sqrt{5}\text{x}+30$ $=\text{x}^2+2\sqrt{5}\text{x}+3\sqrt{5}\text{x}+30$ $=\text{x}(\text{x}+2\sqrt{5})+3\sqrt{5}(\text{x}+2\sqrt{5})$ $=(\text{x}+2\sqrt{5})(\text{x}+3\sqrt{5})$
View full question & answer→Question 772 Marks
Factorise:
$125-8 x^3-27 y^3-90 x y$
Answer$125-8 x^3-27 y^3-90 x y$
$=5^3+(-2 x)^3+(-3 y)^3-3 \times 5 \times(-2 x) \times(-3 y)$
$=[5+(-2 x)+(-3 y)]\left[5^2+(-2 x)^2+(-3 y)^2-5 \times(-2 x)-(-2 x)(-3 y)-5 \times(-3 y)\right]$
$=(5-2 x-3 y)\left(25+4 x^2+9 y^2+10 x-6 x y+15 y\right)$
View full question & answer→Question 782 Marks
Factorise:
$4 x^2+9 y^2+16 z^2+12 x y-24 y z-16 x z$
AnswerWe have:
$4 x^2+9 y^2+16 z^2+12 x y-24 y z-16 x z$
$=(2 x)^2+(3 y)^2+(-4 z)^2+2(2 x)(3 y)+2(3 y)(-4 z)+2(-4 z)(2 x)$
$=[(2 x)+(3 y)+(-4 z)]^2$
$=(2 x+3 y-4 z)^2$
View full question & answer→Question 792 Marks
Find the product.
$(x+y-z)\left(x^2+y^2+z^2-x y+y z+z x\right)$
Answer$(x+y-z)\left(x^2+y^2+z^2-x y+y z+z x\right)$
$=[x+y+(-z)]\left[x^2+y^2+(-z)^2-x y-y \times(-z)-[-z] \times x\right]$
$=x^3+y^3+(-z)^3-3 x \times y \times(-z)$
$=x^3+y^3+(-z)^3-3 x \times y \times(-z)$
$=x^3+y^3-z^3+3 x y z$
View full question & answer→Question 802 Marks
Factorise:
$x^2-(a+b) x+a b$
Answer$x^2-(a+b) x+a b$
$=x^2-a x-b x+a b$
$=x(x-a)-b(x-a)$
$=(x-a)(x-b)$
View full question & answer→Question 812 Marks
Factorise:
$x^2-4 x+3$
Answer$x^2-4 x+3$
$=x^2-3 x-x+3$
$=x(x-3)-1(x-3)$
$=(x-3)(x-1)$
View full question & answer→Question 822 Marks
Factorise: $\frac{2}{3}\text{x}^2-\frac{17}{3}\text{x}-28$
Answer$\frac{2}{3}\text{x}^2-\frac{17}{3}\text{x}-28$ $=\frac{2}{3}\text{x}^2+\frac{7}{3}\text{x}-8\text{x}-28$ $=\frac{\text{x}}{2}(2\text{x}+7)-4(2\text{x}+7)$ $=\Big(\frac{\text{x}}{3}-4\Big)(2\text{x}+7)$
View full question & answer→Question 832 Marks
Factorise:
$125\text{a}^3+\frac{1}{8}$
Answer$125\text{a}^3+\frac{1}{8}$
We know that:
Since $a^2+b^3=(a+b)\left(a^2-a \times b+b^2\right)$
Let us rewrite
$125\text{a}^3+\frac{1}{8}$
$=(5\text{a})^3+\Big(\frac{1}{2}\Big)^3$
$=\Big(5\text{a}+\frac{1}{2}\Big)\bigg[(5\text{a})^2-5\text{a}\times\frac{1}{2}+\Big(\frac{1}{2}\Big)^2\bigg]$
$=\Big(5\text{a}+\frac{1}{2}\Big)\Big(25\text{a}^2-\frac{5\text{a}}{2}+\frac{1}{4}\Big)$
View full question & answer→Question 842 Marks
Factorise:
$x^4-625$
Answer$x^4-625$
$=\left(x^2\right)^2-(25)^2$
$=\left(x^2+25\right)\left(x^2-25\right)$
$=\left(x^2+25\right)\left(x^2-5^2\right)\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
$=\left(x^2+25\right)(x+5)(x-5)$
View full question & answer→Question 852 Marks
Factorise:
$9 a^2+3 a-8 b-64 b^2$
Answer$9 a^2+3 a-8 b-64 b^2$
$=9 a^2-64 b^2+3 a-8 b$
$=(3 a)^2-(8 b)^2+(3 a-8 b)\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
$=(3 a+8 b)(3 a-8 b)+(3 a-8 b)$
$=(3 a-8 b)(3 a+8 b+1)$
View full question & answer→Question 862 Marks
Factorise:
$9-a^2+2 a b-b^2$
Answer$9-a^2+2 a b-b^2$
$=9-\left(a^2-2 a b+b^2\right)$
$=3^2-(a-b)^2\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
$=(3+a-b)(3-a+b)$
View full question & answer→Question 872 Marks
Expand:
$(-3 a+4 b-5 c)^2$
Answer$(-3 a+4 b-5 c)^2=[(-3 a)+(4 b)+(-5 c)]^2$
$=(-3 a)^2+(4 b)^2+(-5 c)^2+2(-3 a)(4 b)+2(4 b)(-5 c)+2(-3 a)(-5 c)$
$=9 a^2+16 b^2+25 c^2-24 a b-40 b c+30 a c$
View full question & answer→Question 882 Marks
Factorise:
$a^{12}-b^{12}$
Answer$a^{12}-b^{12}$
$=\left(a^6\right)^2-\left(b^6\right)^2$
$=\left(a^6-b^6\right)\left(a^6+b^6\right)$
$=\left[\left(a^3\right)^2-\left(b^3\right)^2\right]\left[\left(a^2\right)^3+\left(b^2\right)^3\right]$
$=\left(a^3-b^3\right)\left(a^3+b^3\right)\left[\left(a^2+b^2\right)\left(a^4-a^2 b^2+b^4\right)\right]$
$=(a-b)\left(a^2+a b+b^2\right)(a+b)\left(a^2-a b+b^2\right)\left(a^2+b^2\right)\left(a^4-a^2 b^2+b^4\right)$
$=(a-b)(a+b)\left(a^2+b^2\right)\left(a^2+a b+b^2\right)\left(a^2-a b+b^2\right)\left(a^4-a^2 b^2+b^4\right)$
View full question & answer→Question 892 Marks
Evaluate:
$(99)^3$
Answer$(99)^3$
$=(100-1)^3$
$=(100)^3-(1)^3-3(100)^2 \times(1)+3(100)(1)^2$
$=1000000-1-30000+300$
$=1000300-30001$
$=970299$
View full question & answer→Question 902 Marks
Factorise:
$24 x^2-41 x+12$
Answer$24 x^2-41 x+12$
$=24 x^2-32 x-9 x+12$
$=8 x(3 x-4)-3(3 x-4)$
$=(3 x-4)(8 x-3)$
View full question & answer→Question 912 Marks
Factorise:
$27 x^3-y^3-z^3-9 x y z$
Answer$27 x^3-y^3-z^3-9 x y z$
$=(3 x)^3-y^3-z^3-3 \times(3 x) \times(-y) \times(-z)$
We know,
$a^3+b^3+c^3-3 a b c$
$=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
$a=3 x, b=-y, c=-z$
$(3 x)^3-y^3-z^3-3 \times(3 x) \times(-y) \times(-z)$
$=(3 x-y-z)\left(9 x^2+y^2+z^2+3 x y-y z+3 x z\right)$
View full question & answer→Question 922 Marks
Evaluate:
$(103)^3$
Answer$(103)^3$
$=(100+3)^3$
$=(100)^3+(3)^3+3(100)^2 \times(3)+3(100)(3)^2$
$=1000000+27+90000+2700$
$=1092727$
View full question & answer→Question 932 Marks
Factorise:
$x^9-y^9$
Answer$x^9-y^9$
$=\left(x^3\right)^3-\left(y^3\right)^3$
$=\left[\left(x^3-y^3\right)\right]\left[\left(x^3\right)^2+x^3 y^3+\left(y^3\right)^2\right]$
$=\left[(x-y)\left(x^2+x y+y^2\right)\left(x^6+x^3 y^3+y^6\right)\right.$
View full question & answer→Question 942 Marks
Factorise:
$(a-b)^3+(b-c)^3+(c-a)^3$
Answer$(a-b)^3+(b-c)^3+(c-a)^3$
$\text { Putting }(a-b)=x,(b-c)=y \text { and }(c-a)=z$
$\text { We get: }(a-b)^3+(b-c)^3+(c-a)^3$
$=x^3+y^3+z^3[(x+y+z)=(a-b)+(b-c)+(c-a)=0]$
$=3 x y z\left[(x+y+z)=0 \Rightarrow x^3+y^3+z^3=3 x y z\right]$
$=3(a-b)(b-c)(c-a)$
View full question & answer→Question 952 Marks
Factorise:
$(a x+b y)^2+(b x-a y)^2$
Answer$(a x+b y)^2+(b x-a y)^2$
$=a^2 x^2+b^2 y^2+2 a b x y+b^2 x^2+a^2 y^2-2 a b x y$
$=a^2 x^2+b^2 y^2+b^2 x^2+a^2 y^2$
$=a^2 x^2+b^2 x^2+b^2 y^2+a^2 y^2$
$=x^2\left(a^2+b^2\right)+y^2\left(a^2+b^2\right)$
$=\left(a^2+b^2\right)\left(x^2+y^2\right)$
View full question & answer→Question 962 Marks
Factorise: $\text{x}^2+3\sqrt{3}\text{x}+6$
Answer$\text{x}^2+3\sqrt{3}\text{x}+6$ $=\text{x}^2+2\sqrt{3}\text{x}+\sqrt{3}\text{x}+6$ $=\text{x}(\text{x}+2\sqrt{3})+\sqrt{3}(\text{x}+2\sqrt{3})$ $=(\text{x}+2\sqrt{3})(\text{x}+\sqrt{3})$
View full question & answer→Question 972 Marks
Factorise:
$a^3-0.064$
Answer$a^3-0.064$
$=(a)^3-(0.4)^3$
$=(a-0.4)\left[(a)^2+a \times 0.4+(0.4)^2\right] \text { Since } a^3-b^3=(a-b)\left(a^2+a \times b+b^2\right)$
$=(a-0.4)\left(a^2+0.4 a+0.16\right)$
View full question & answer→Question 982 Marks
Factorise:
$81 x^4-y^4$
Answer$81 x^4-y^4$
$=\left(9 x^2\right)^2-\left(y^2\right)^2$
$=\left(9 x^2-y^2\right)\left(9 x^2+y^2\right)$
$=\left[(3 x)^2-y^2\right]\left(9 x^2+y^2\right)$
$=(3 x-y)(3 x+y)\left(9 x^2+y^2\right)$
View full question & answer→Question 992 Marks
Factorise:
$8 a^3+125 b^3-64 c^3+120 a b c$
Answer$8 a^3+125 b^3-64 c^3+120 a b c$
$=(2 a)^3+(5 b)^3+(-4 c)^3-3 \times(2 a) \times(5 b) \times(-4 c)$
$=(2 a+5 b-4 c)\left[(2 a)^2+(5 b)^2+(-4 c)^2-(2 a)(5 b)-(5 b)(-4 c)-(2 a) \times(-4 c)\right]$
$=(2 a+5 b-4 c)\left(4 a^2+25 b^2+16 c^2-10 a b+20 b c+8 a c\right)$
View full question & answer→Question 1002 Marks
Factorise:
$(x+2)^3+(x-2)^3$
Answer$(x+2)^3+(x-2)^3$
$=[(x+2)+(x-2)]\left[(x+2)^2-(x+2)(x-2)+(x-2)^2\right]$
$=(2 x)\left(x^2+4 x+4-x^2+4+x^2-4 x+4\right)$
$=2 x\left(x^2+12\right)$
View full question & answer→