Question

without calculating, find the value of​

Answer 1

$\begin{aligned}&125(x-y)^3+(5y-3z)^3+(3z-5x)^3\\ \\ \Longrightarrow\ \ &5^3(x-y)^3+(5y-3z)^3+(3z-5x)^3\\ \\ \Longrightarrow\ \ &(5(x-y))^3+(5y-3z)^3+(3z-5x)^3\\ \\ \Longrightarrow\ \ &(5x-5y)^3+(5y-3z)^3+(3z-5x)^3\end{aligned}$

Now, let,

p = 5x - 5y

q = 5y - 3z

r = 3z - 5x

we can see that,

$p+q+r=(a-b)+(b-c)+(c-a)=a-b+b-c+c-a=0$

So, let me recall the given identity.

$\begin{aligned} &p^3+q^3+r^3-3pqr=(p+q+r)(p^2+q^2+r^2-pq-qr-rp)\\ \\ \Longrightarrow\ \ &p^3+q^3+r^3-3pqr=0\ \ \ [\because p+q+r=0]\\ \\ \Longrightarrow\ \ &p^3+q^3+r^3=3pqr\\ \\ \Longrightarrow\ \ &(5x-5y)^3+(5y-3z)^3+(3z-5x)^3=3(5x-5y)(5y-3z)(3z-5x)\end{aligned}$

Here we get factorized form of the equation given in the question. It can be expanded if necessary.

Hence found out without calculating cubes!