Question

factorise (3a-5b)^3+(4c-3a)^3+(5b-4c)^3​

Answer 1

Answer:

The answer is $3(3a-5b)(4c-3a)(5b-4c)$

Step-by-step explanation:

Let $3a-5b=x$, $4c-3a=y$ and $5b-4c=z$

   $\Rightarrow x+y+z=3a-5b+4c-3a+5b-4c=0$

Thus

$(3a-5b)^3+(4c-3a)^3+(5b-4c)^3\\=x^3+y^3+z^3\\=3xyz=3(3a-5b)(4c-3a)(5b-4c)$

( using $x+y+z=0\Rightarrow x^2+y^3+z^3=3xyz$ )