1.

Let x, y, z be non-zero real numbers such that x/y+y/z+z/x=7 and y/x+z/y+x/z=9, then x3/y3+y3/z3+z3/x3-3 is equal to (A) 152 (B) 153 (D) 154 (D) 155

Answer»

Correct option(c)

Explanation:

a3 + b3 + c3-3abc = [a + b + c] [(a + b + c)2 - 3(ab + bc + ca)] 

= [7] [(7)2 - 3(9)] = 7(49 - 27) = 7 × 22 = 154



Discussion

No Comment Found

Related InterviewSolutions