Question

x^6-11x^3+30/x^3-6

Answer 1

here, first substitue x^3=t
u hav a factorisable quadratic equation in the numerator.
it factors to : (t+5)(t-6) and so (t-6) get cancelled...u hav t+5 i.e. x^3+5

Answer 2

The answer is: (x^3-5)

Answer 3

Not x^5, brackett went wrong somewhere while computing

Answer 4

not x^3+5*

Answer 5

oops sorry it is x^3-5 only.....factorisation error :)

Answer 6

The reason my answer is correct is because the top part or numerator factors to (x^3-6)(x^3-5), thus the (x^3-6) in the numerator and denominator cancel, leaving us with (x^3-5)