Question

Factorize 2√2a³+3√3b³+c³-2√6abc

Answer 1

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 )