Question

without finding cubes, factorise:
1.(2r-3s)^3+(3s-5t)^3+(5t-2r)^3

Answer 1

Heya,

Let,
a = (2r-3s)
b = (3s-5t)
c = (5t-2r)

If a + b + c = 0

Then a^3 + b^3 + c^3 = 3abc

(2r-3s)^3+(3s-5t)^3+(5t-2r)^3 = 3 × (2r-3s) × (3s-5t) × (5t-2r)

=> 3 × (+ 6rs - 10rt + 10rt - 4r^2 - 9s^2 + 15st - 15st + 6sr -25t^2)

=> 3 × ( +12sr - 4r^2 - 9s^2 - 25t^2)

=> 36sr - 12r^2 - 27s^2 - 75t^2

Hope my answer helps you :)

Regards,
Shobana

Answer 2

this is given picture is the answer of your question.