factorise 2root2a^3-8b^3-27c^3+8root abc
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=(√2a)^3 +(-2b)^3 + (-3c)^3 - 3(√2a×-2b×-3c)
=using identity a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2 +c^2-ab-bc-ca)
=(√2a-2b-3c){(√2a)^2+(-2b)^2+(-3c)^2-(√2a)(-2b) -(-2b)(-3c)- (-3c)(√2a)
=(√2a-2b-3c)(2a^2+4b^2+3c^2+2√2ab-6bc+3√2ac)
=using identity a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2 +c^2-ab-bc-ca)
=(√2a-2b-3c){(√2a)^2+(-2b)^2+(-3c)^2-(√2a)(-2b) -(-2b)(-3c)- (-3c)(√2a)
=(√2a-2b-3c)(2a^2+4b^2+3c^2+2√2ab-6bc+3√2ac)
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