Question

a cube minus 1 by a ​

Answer 1

We have:

$\dfrac{a^3-1}{a}$

We have to find, the factorisation of $\dfrac{a^3-1}{a}$ is:

Solution:

∴ $\dfrac{a^3-1}{a}$

$=\dfrac{a^3-1^3}{a}$

Using the algebraic identity:

$x^{3} -y^{3} =(x-y)(x^{2} +xy+y^{2} )$

$=\dfrac{(a-1)(a^2+a+1)}{a}$

Thus, the factorisation of $\dfrac{a^3-1}{a}$ is $\dfrac{(a-1)(a^2+a+1)}{a}$.