Maths M.C.Q

$\Big(7\text{a}+\frac{1}{2}\Big)\Big(7\text{a}-\frac{1}{2}\Big)=(7\text{a})^2-\Big(\frac{1}{2}\Big)^2$
[by using identity $(a+b)(a-b)=a^2-b^2$]
$\Rightarrow\Big(7\text{a}+\frac{1}{2}\Big)\Big(7\text{a}-\frac{1}{2}\Big)=49\text{a}^2-\frac{1}{4}$
$\Rightarrow49\text{a}^2-\text{b}=49\text{a}^2-\frac{1}{4}$
$\Rightarrow\text{b}=\frac{1}{4}$
Hence, correct option is $(b)$.

$a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
If $\mathrm{a}+\mathrm{b}+\mathrm{c}=0$, then
$a^3+b^3+c^3-3 a b c=0$
$\Rightarrow a^3+b^3+c^3=3 a b c$
Now, consider $\frac{\mathrm{a}^2}{\mathrm{bc}}+\frac{\mathrm{b}^2}{\mathrm{ca}}+\frac{\mathrm{c}^2}{\mathrm{ab}}$
Multiplying dividing by a. b. and c in $\frac{\mathrm{a}^2}{\mathrm{bc}} \cdot \frac{\mathrm{b}^2}{\mathrm{ca}}$ and $\frac{\mathrm{c}^2}{\mathrm{ab}}$ respectively. we get
$\frac{\mathrm{a}^3}{\mathrm{abc}}+\frac{\mathrm{b}^3}{\mathrm{bca}}+\frac{\mathrm{c}^3}{\mathrm{cab}}$
$=\frac{\mathrm{a}^3+\mathrm{b}^3+\mathrm{c}^3}{\mathrm{abc}}$
$=\frac{3 \mathrm{abc}}{\mathrm{abc}} \ldots[\text { From (1)] }$
$=3$
Hence, correct option is $(d)$.

$a-b=-8$
$(a-b)^2=64$
$a^2+b^2-2 a b=64$
$a^2+b^2-2 a b+3 a b=64+3 a b$
$a^2+b^2+a b=64+3(-12)$
$a^2+b^2+a b=64-36$
$a^2+b^2+a b=28$
Now
$a^3-b^3=(a-b)\left(a^2+b^2+a b\right)$
$=(-8)(28)$
$=-224$
Hence, correct option is $(c)$.

Given expression is $\left(x^2-1\right)\left(x^4+x^2+1\right)$
Let $\mathrm{x}^2=\mathrm{A}$ and $1=\mathrm{B}$
Then, we have
$(A-B)\left(A^2+A B+B^2\right)$
$=A^3-B^3$
$=\left(X^2\right)^3-(1)^3$
$=X^6-1$
Hence, correct option is $(c)$.

$(x-y)(x+y)=x^2-y^2\left[\text { by identity }(a+b)(a-b)=a^2-b^2\right]$
$\left(x^2-y^2\right)\left(x^2+y^2\right)=x^4-y^4$
$\left(x^4-y^4\right)\left(x^4+y^4\right)=x^8-y^8$
Now,
$(x-y)(x+y)\left(x^2+y^2\right)\left(x^4+y^4\right)$
$=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)$
$=\left(x^4-y^4\right)\left(x^4+y^4\right)$
$=x^8-y^8$
Hence, correct option is $(b)$.

$\text { If } a+b+c=0 \text { then, } a^3+b^3+c^3=3 a b c$
$\text { Now, }\left(a^2-b^2\right)+\left(b^2-c^2\right)+\left(c^2-a^2\right)=a^2-b^2+b^2-c^2+c^2-a^2=0$
$\Rightarrow\left(a^2-b^2\right)^3+\left(b^2-c^2\right)^3+\left(c^2-a^2\right)^3=3\left(a^2-b^2\right)\left(b^2-c^2\right)\left(c^2-a^2\right)$
$\text { Again, }(a-b)+(b-c)+(c-a)=a-b+b-c+c-a=0$
$\Rightarrow(a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)$
Thus, we have
$\frac{(\text{a}^2-\text{b}^2)^3+(\text{b}^2-\text{c}^2)^3+(\text{c}^2-\text{a}^2)^3}{(\text{a}-\text{b})^3+(\text{b}-\text{c})^3+(\text{c}-\text{a})^3}$
$=\frac{3(\text{a}^2-\text{b}^2)(\text{b}^2-\text{c}^2)(\text{c}^2-\text{a}^2)}{3(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$=\frac{(\text{a}-\text{b})(\text{a}+\text{b})(\text{b}-\text{c})(\text{b}+\text{c})(\text{c}-\text{a})(\text{c}+\text{a})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$=(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})$
Hence, correct option is $(d)$.

Given, $a+b+c=9$
Hence, $(a+b+c)^2=81$
So, $a^2+b^2+c^2+2 a b+2 b c+2 c a=81$
i.e. $a^2+b^2+c^2+2(a b+b c+c a)=81$
i.e. $a^2+b^2+c^2+2(23)=81$
i.e. $a^2+b^2+c^2=81-46=35$
Now, $a^3+b^3+c^3-3 a b c$
$=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
$=(a+b+c)\left[\left(a^2+b^2+c^2\right)-(a b+b c+c a)\right]$
$=(9)[35-23]$
$=9 \times 12$
$=108$
Hence, correct option is $(a)$.

Given expression is $75 \times 75+2 \times 75 \times 25+25 \times 25$
Let $75=\mathrm{a}$ and $25=\mathrm{b}$
Then, we have
$a \times a+2 \times a \times b+b \times b$
$=a^2+2 a b+b^2$
$=(a+b)^2$
$=(75+25)^2$
$=(100)^2$
$=10000$
Hence, correct option is $(a)$.

$\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}=-1$
$\Rightarrow\frac{\text{a}^2+\text{b}^2}{\text{ab}}=-1$
$\Rightarrow\text{a}^2+\text{b}^2+\text{ab}=0$
Now using identity
$a^3-b^3$
$=(a-b)\left(a^2+b^2+a b\right)$
$=(a-b)(0)\left(\because a^2+b^2+a b=0\right)$
$=0$
Hence, correct option is $(d)$.

$\Big(\text{x}-\frac{1}{\text{x}}\Big)^3=\text{x}^3-\frac{1}{\text{x}^3}-3\not\text{x}\frac{1}{\not\text{x}}\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\text{x}^3-\frac{1}{\text{x}^3}=\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)-\text{x}^3-\frac{1}{\text{x}^3}=0$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)-14=0$
Let $\Rightarrow\text{x}-\frac{1}{\text{x}}=\text{t}$
$\Rightarrow \mathrm{t}^3+3 \mathrm{t}-14=0$
$\Rightarrow \mathrm{t}^3-2 \mathrm{t}^2+2 \mathrm{t}^2-4 \mathrm{t}+7 \mathrm{t}-14=0$
$\Rightarrow \mathrm{t}(\mathrm{t}-2)+2 \mathrm{t}(\mathrm{t}-2)+7(\mathrm{t}-2)=0$
$\Rightarrow(\mathrm{t}-2)(\mathrm{t}+2 \mathrm{t}+7)=0$
$\Rightarrow \mathrm{t}^2+2 \mathrm{t}+7=0 \text { has no real roots }$
So, $t=2$ is a solution
$\Rightarrow\text{x}-\frac{1}{\text{x}}=2$
Hence, correct option is $(d)$.

Let $\text{a}^{\frac{1}{3}}=\text{A},\ \text{b}^{\frac{1}{3}}=\text{B}$ and $\text{c}^{\frac{1}{3}}=\text{C}$
Now, $\mathrm{A}+\mathrm{B}+\mathrm{C}=0$ (given)
If $A+B+C=0$, then $A^3+B^3+C^3-3 A B C=0$
$\Rightarrow A^3+B^3+C^3-3 A B C=0$
$\Rightarrow A^3+B^3+C^3=3 A B C \ldots(1)$
$\begin{Bmatrix}\text{A}=\text{a}^{\frac{1}{3}},\ \text{B}=\text{b}^{\frac{1}{3}},\ \text{C}=\text{c}^{\frac{1}{3}}\\\text{A}^3=\text{a},\ \text{B}^3=\text{b},\ \text{C}^3=\text{c}\end{Bmatrix}$
Then, equation (1) becomes
$\text{a}+\text{b}+\text{c}=3(\text{abc})^{\frac{1}{3}}$
Cubing both Sides of above equation, we get
$(a+b+c)^3=27 a b c$
Hence, correct option is $(b)$.

By using identity $(a+b)^2=a^2+b^2+2 a b$.
we have,
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\Big(\frac{1}{\text{x}}\Big)^2+2\times\not\text{x}\times\frac{1}{\not\text{x}}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow(5)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$ $\Big\{\text{x}+\frac{1}{\text{x}}=5\text{ given}\Big\}$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=25-2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=23$
Hence, correct option is $(c)$.

We know that $(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
Here, $a+b+c=9, a b+b c+c a=23$
Thus, we have
$(9)^2=a^2+b^2+c^2+2(23)$
$81=a^2+b^2+c^2+46$
$a^2+b^2+c^2=81-46$
$a^2+b^2+c^2=35$
Hence, correct option is $(a)$.

$(a+b)(a-b)\left(a^2-a b+b^2\right)\left(a^2+a b+b^2\right)$
$=\left(a^2-b^2\right)\left(a^2+b^2-a b\right)\left(a^2+b^2-a b\right)$
$=\left(a^2-b^2\right)\left\{\left(a^2+b^2\right)^2-(a b)^2\right\}$
$=\left(a^2-b^2\right)\left\{a^4+b^4+2 a^2 b^2-a^2 b^2\right\}$
$=\left(a^2-b^2\right)\left\{a^4+b^4+a^2 b^2\right\}$
$=\left\{a^6+a^2 b^4+a^4 b^2-b^2 a^4-b^6-b^4 a^2\right\}$
$=a^6-b^6$
Hence, correct option is $(b)$.