Q. 10 B5.0( 1 Vote )

# Verify that:

27a^{3} + b^{3} + c^{3} – 9abc = (3a + b + c)[3a^{2} + b^{2} + c^{2} – 3ab – bc – 3ac]

Answer :

Taking L.H.S

27a^{3} + b^{3} + c^{3} – 9abc = (3a)^{3} + b^{3} + c^{3} – 3(3a)(b)(c)

Using identity:

x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z)(x^{2} + y^{2} + z^{2 –} xy – yz – zx)

⇒ 27a^{3} + b^{3} + c^{3} – 9abc = (3a + b + c)((3a)^{2} + b^{2} + c^{2} – 3ab – bc – c(3a))

= (3a + b + c)[3a^{2} + b^{2} + c^{2} – 3ab – bc – 3ac]

= R.H.S

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