Q11 of 90 Page 33

If x + y + z = 0 then verify that x3 + y3 + z3 = 3xyz.

Given that,

x + y + z = 0…..(1)


And, we have proved that -



Now, substituting (1) in (2), we get –


x3 + y3 + z3 – 3xyz = 0


⇒ x3 + y3 + z3 = 3xyz


More from this chapter

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10

Verify that:

x3 + y3 + z3 – 3xyz = 1/2 (x + y + z)[(x – y)2 + (y – z)2 + (z – x)2]

 10

Verify that:

27a3 + b3 + c3 – 9abc = (3a + b + c)[3a2 + b2 + c2 – 3ab – bc – 3ac]

 12

Calculate using appropriate algebraic identity:

(30)3 + (20)3 + (– 50)3

 12

Calculate using appropriate algebraic identity:

(– 15)3 + (28)3 + (– 13)3


[Hint: Use the identity : If x + y + z = 0 the x3 + y3 + z3 = 3xyz]