factorise. 125-2√2a³+3√3b³+15√6ab
Answers
it is given that , 125 - 2√2a³ + 3√3b³ + 15√6ab
or, (5)³ + (-√2 a)³ + (√3 b)³ - 3(5)(-√2a)(√3b)
we know, from Algebraic identity,
x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)
if we assume, 5 =x, -√2a = y and √3b = z
then, (5)³ + (-√2 a)³ + (√3 b)³ - 3(5)(-√2a)(√3b) = x³ + y³ + z³ + 3xyz
so, (5)³ + (-√2 a)³ + (√3 b)³ - 3(5)(-√2a)(√3b) = {5 + (-√2a) + (√3b)}{(5)² + (-√2a)² + (√3b)² - 5(-√2a) - (-√2a)(√3b) - (√3b)(5)}
= (5 - √2a + √3b)(25 + 2a² + 3b² + 5√2a +√6ab - 5√3b)
hence, factorisation of 125-2√2a³+3√3b³+15√6ab = (5 - √2a + √3b)(25 + 2a² + 3b² + 5√2a +√6ab - 5√3b)
Answer:
Step-by-step explanation:
Our aim is to factorize 125-2√2a³+3√3b³+15√6ab.
We need to consider some aspects in order to compute this factorization.
Firstly we need to know, from algebraic identity that:
- x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)
In our case we need to make some substitutions, so let us consider 5 = x, -√2a = y and √3b = z, then we have:
- (5)³ + (-√2 a)³ + (√3 b)³ - 3(5)(-√2a)(√3b) = x³ + y³ + z³ + 3xyz
- = (5 - √2a + √3b)(25 + 2a² + 3b² + 5√2a +√6ab - 5√3b)
Hence, our factorization of 125-2√2a³+3√3b³+15√6ab is (5 - √2a + √3b)(25 + 2a² + 3b² + 5√2a +√6ab - 5√3b)
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