Q. 10 B5.0( 1 Vote )

# 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

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