Q10 of 90 Page 33

Verify that:

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

Taking L.H.S

27a3 + b3 + c3 – 9abc = (3a)3 + b3 + c3 – 3(3a)(b)(c)


Using identity:


x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)


⇒ 27a3 + b3 + c3 – 9abc = (3a + b + c)((3a)2 + b2 + c2 – 3ab – bc – c(3a))


= (3a + b + c)[3a2 + b2 + c2 – 3ab – bc – 3ac]


= R.H.S


More from this chapter

All 90 →
9

Factorize:

(i) 64a3 + 27b3


(ii) 125x3 – 8y3

 10

Verify that:

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

 11

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

 12

Calculate using appropriate algebraic identity:

(30)3 + (20)3 + (– 50)3