factorise 2✓2a^2+3✓3b^3+✓5(5-3✓6ab)
please give step by step explanation.
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Question should be 2√2a³ + 3√3b³ + c³ - 3√6abc
= √2a × √2a × √2a + √3b × √3b × √3b + c × c × c - 3(√2a)(√3b)(c)
= (√2a)³ + (√3b)² + c³ - 3(√2a)(√3b)(c)
We know, x³ + y³ + z³ - 3xyz = 1/2(x + y + z) {x² + y² + z² - xy - yz - yz}
Use it here,
Then, (√2a)³ + (√3b)² + c³ - 3(√2a)(√3b)(c) = 1/2(√2a + √3b + c){(√2a)² + (√3b)² + c - (√2a)(√3b) - (√3b)(c) - (c)(√2a)}
= 1/2(√2a + √3b + c)(2a² + 3b² + c² - √6ab - √3bc - √2ca )
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