Before answering let me make you familiar with an algebraic identity

a³+b³+c³-3abc= (a+b+c)(a²+b²+c²-ab-bc-ca)

In this case, (a+b+c) = 2r-3s+3s-4t+4t-2r

⇒0

So, (2r-3s)³+(3s-4t)³+(4t-2r)³ - 3(2r-3s)(3s-4t)(4t-2r) = 0

⇒(2r-3s)³+(3s-4t)³+(4t-2r)³ = 3(2r-3s)(3s-4t)(4t-2r)

And that's it.

Answer:

3(2r-3s)(3s-4t)(4t-2r)

Step-by-step explanation:

identity :- [a3 + b3 +c3 -3abc = (a+b+c) (a2 + b2 +c2 - ab -bc - ca]

(a+b+c) = 0 [In this case only]

rhs = 0

a3 + b3 +c3 -3abc =0

a3 + b3 +c3 = 3abc

and we get 3(2r-3s)(3s-4t)(4t-2r)