The statement is asking us to prove that if x = 2 then x^3 - x^2 - x = 2 and if x^3 - x^2 - x - 2 = 0 then x = 2 is the only real solution. Easy to show that 2^3 - 2^2 - 2 is equal to 2 . Therefore, we have shown that if x = 2 , then x^3 - x^2 - x = 2 . To prove the converse, we have to show that x = 2 is the only real zero of x^3 - x^2 - x -2 . To find the other zeros, we divide x^3 - x^2 - x - 2 by x - 2 . We have (x^3 - x^2 - x - 2) \div (x-2) = x^2 + x + 1 . Note that neither of the zeros \frac{-1 \pm \sqrt{3} i}{2} of x^2 + x + 1 obtained using the quadratic formula is real, leaving x = 2 as the only real solution. Hence, we have shown that if x is real and x^3 - x^2 - x = 2 then x has to be equal to 2 .