Question

Factorise $125a^3+b^3+64c^3-60abc$ using the identity$a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)[{(a-b)}^2+{(b-c)}^2+{(c-a)}^2]$

Answer 1

The given identity is $a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]$ ....... $\left(i\right)$

We can write $125a^3$ as $(5a{)}^3$ and $64c^3$ as $(4c{)}^3$

Using the above identity taking $5a=x$, $b=y$ and $4c=z$, the equation $125a^3+b^3+64c^3-60abc$ can be factorised as follows:

$x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)\left[\right(x-y{)}^2+(y-z{)}^2+(z-x{)}^2]$

$=\frac{1}{2}(5a+b+4c)\left[\right(5a-b{)}^2+(b-4c{)}^2+(4c-5a{)}^2]$

Hence, $125a^3+b^3+64c^3-60abc=\frac{1}{2}(5a+b+4c)(50a^2+2b^2+32c^2-10ab-8bc-40ca)$