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2√2a3 +3√3b3+c3-3√6abc

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2√2a3 +3√3b3+c3-3√6abc

Posted by Ishaan Sood 5 years, 4 months ago

CBSE > Class 09 > Mathematics

1 answers

Ankit Raj 5 years, 4 months ago

Solution:- since, x=√2a y=√3b z=c So, x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx) Now, (√2a)³+(√3b)³+(c)³-3•√2a•√3b•c =(√2a+√3b+c)(2a²+3b²+c²-√6ab-√3bc-√2ca).

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CBSE > Class 09 > Mathematics