Question
Factorize the polynomial:
$8 a^3-b^3-12 a^2 b+6 a b^2$
$8 a^3-b^3-12 a^2 b+6 a b^2$
✓
Answer
$8 a^3-b^3-12 a^2 b+6 a b^2$
The expression $8 a^3-b^3-12 a^2 b+6 a b^2$ can also be written as $=(2 a)^3-(b)^3-3 \times 2 a \times 2 a \times b+3 \times 2 a \times b \times b$ $=(2 a)^3-(b)^3-3 \times 2 a \times b(2 a-b)$.
Using identity $(x-y)^3=x^3-y^3-3 x y(x-y)$ with respect to the expression $(2 a)^3-(b)^3-3 \times 2 a \times b(2 a-b)$, we get $(2 a-b)^3$
Therefore, after factorizing the expression $8 a^3-b^3-12 a^2 b+6 a b^2$, weget $(2 a-b)^3$
The expression $8 a^3-b^3-12 a^2 b+6 a b^2$ can also be written as $=(2 a)^3-(b)^3-3 \times 2 a \times 2 a \times b+3 \times 2 a \times b \times b$ $=(2 a)^3-(b)^3-3 \times 2 a \times b(2 a-b)$.
Using identity $(x-y)^3=x^3-y^3-3 x y(x-y)$ with respect to the expression $(2 a)^3-(b)^3-3 \times 2 a \times b(2 a-b)$, we get $(2 a-b)^3$
Therefore, after factorizing the expression $8 a^3-b^3-12 a^2 b+6 a b^2$, weget $(2 a-b)^3$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
It being given that $\sqrt{3}=1.732,\sqrt{5}=2.236,\sqrt{6}=2.449$ and $\sqrt{10}=3.162,$ find to three places of decimal, the value of the following:
$\frac{1}{4\sqrt{3}-3\sqrt{5}}$
→$\frac{1}{4\sqrt{3}-3\sqrt{5}}$
If the angles of a triangle are in the ratio $1 : 2 : 3$, determine three angles.
→$ABCD$ is a quadrilaeter in which $AB = BC$ and $AD = CD$. Show that $BD$ bisects both the angles $ABC$ and $ADC.$
→$ABCD$ is a parallelogram in which $AB = 9.5\ cm$ and its perimeter is $30\ cm.$ Find the length of each side of the parallelogram.
→In the adjoining figure, $\text{MNPQ}$ and $\text{ABPQ}$ are parallelograms and $T$ is any point on the side $BP.$ Prove that
$i. \text{ar}(\text{MNPQ})=\text{ar}(\text{ABPQ})$
$ii. \text{ar}(\triangle\text{ATQ})=\frac{1}{2}\text{ar}\text{(MNPQ}).$
→$i. \text{ar}(\text{MNPQ})=\text{ar}(\text{ABPQ})$
$ii. \text{ar}(\triangle\text{ATQ})=\frac{1}{2}\text{ar}\text{(MNPQ}).$
If the radius of the base of a cone is halved, keeping the height same, what is the ratio of the volume of the reduced cone to that of the original cone?
Draw a histogram for the daily earnings of 30 drug stores in the following table:
| Daily earnings (in ₹ ): | 450-500 | 500-550 | 550-600 | 600-650 | 650-700 |
| Number of Stores: | 16 | 10 | 7 | 3 | 1 |
Factorize:
$9(2 a-b)^2-4(2 a-b)-13$
→$9(2 a-b)^2-4(2 a-b)-13$
If both $x -2$ and $x -\frac{1}{2}$ are factors of $px ^2+5 x + r$, show that $p = r$.
→