Factorise :
$1+64 x^{3}$
Factorise :
$1+64 x^{3}$
We have,
$1+64 x^{3}=(1)^{3}+(4 x)^{3}$
$=(1+4 x)\left\{(1)^{2}-(1)(4 x)+(4 x)^{2}\right\}$ $\left[\because a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$
$=(1+4 x)\left(1-4 x+16 x^{2}\right)$
$=(1+4 x)\left(16 x^{2}-4 x+1\right)$
$=(4 x+1)\left(16 x^{2}-4 x+1\right)$
Similar Questions
Find the zero of the polynomial in each of the following cases
$q(y)=\pi y+3.14$
Find the zero of the polynomial in each of the following cases
$q(y)=\pi y+3.14$
Factorise the following quadratic polynomials by splitting the middle term
$15 x^{2}+7 x-2$
Factorise the following quadratic polynomials by splitting the middle term
$15 x^{2}+7 x-2$
Find the quotient and the remainder when $2 x^{2}-7 x-15$ is divided by
$2 x+1$
Find the quotient and the remainder when $2 x^{2}-7 x-15$ is divided by
$2 x+1$
Expand the following:
$\left(\frac{1}{x}+\frac{y}{3}\right)^{3}$
Expand the following:
$\left(\frac{1}{x}+\frac{y}{3}\right)^{3}$
Without finding the cubes, factorise
$(x-2 y)^{3}+(2 y-3 z)^{3}+(3 z-x)^{3}$
Without finding the cubes, factorise
$(x-2 y)^{3}+(2 y-3 z)^{3}+(3 z-x)^{3}$