Question
Factorise : $\mathrm{a}^3-\frac{27}{a^3}$
✓
Answer
$ a^3-\frac{27}{a^3}$
$ =(a)^3-\left(\frac{3}{a}\right)^3$
$ =\left(a-\frac{3}{a}\right)\left[a^2+a \times \frac{3}{a}+\left(\frac{3}{a}\right)^2\right]$
$\left[\because a^3+b^3=(a b)\left(a^2\right.\right. \left.\left.+a b+b^2\right)\right]$
$ =\left(a-\frac{3}{a}\right)\left(a^2+3+\frac{9}{a^2}\right)$
$ =(a)^3-\left(\frac{3}{a}\right)^3$
$ =\left(a-\frac{3}{a}\right)\left[a^2+a \times \frac{3}{a}+\left(\frac{3}{a}\right)^2\right]$
$\left[\because a^3+b^3=(a b)\left(a^2\right.\right. \left.\left.+a b+b^2\right)\right]$
$ =\left(a-\frac{3}{a}\right)\left(a^2+3+\frac{9}{a^2}\right)$
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
The base $BC$ of $\triangle ABC$ is divided at $D$ .so that $BD=\frac{1}{2} DC$.Prove that area of $ΔABD=\frac{1}{3}$ of the area of $ΔABC.$
→Factorise : $1 - 2a - 2b - 3 (a + b)^2$
→Solve the equations:
If $2 x+y=32$ and $3 x+4 y=68$, find the value of $\frac{x}{y}$.
→If $2 x+y=32$ and $3 x+4 y=68$, find the value of $\frac{x}{y}$.
If $a + b = 11$ and $a^2+ b^2= 65;$ find $a^3+ b^3.$
→In the given figure, $AB > AC$. If BO and CO are the bisectors of $\angle B$ and $\angle C$ respectively, prove that $BO > CO$.
→Find the amount and the compound interest on ₹ 10000 for 1 year at 12% per annum, compounded half-yearly.
Evaluate : $\frac{3 \sin 3 A+2 \cos \left(2 A+5^{\circ}\right)}{2 \cos 3 A-\sin \left(2 A-10^{\circ}\right)}$, when $A =20$.
→Find the co$-$ordinates of points on the $x-$axis which are at a distance of $17\ units$ from the point $(11, -8).$
→Using $(a-b)^2=\left(a^2+b^2-2 a b\right)$, evaluate :$(992)^2$
→Solve the pair of linear $($simultaneous$)$ equation by the method of elimination by substitution$:6x = 7y + 7,7y - x = 8$
→