An algebra problem by Daman Deep singh

Algebra Level 3

If x = 2 + 2 1 3 + 2 2 3 \large x=2+2^\frac 13 +2^\frac 23 , then find the value of x 3 − 6 x 2 + 6 x \large x^3-6x^2+6x .

0 4 6 2

Daman Deep Singh by Daman Deep singh , 22, India

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2 solutions

Rishabh Jain
Jun 18, 2016

Use ( a + b ) 3 = a 3 + b 3 + 3 a b ( a + b ) \small{\color{grey}{\text{Use } (a+b)^3=a^3+b^3+3ab(a+b)}}

( x − 2 ) 3 ⏟ x 3 − 6 x 2 + 12 x − 8 = ( 2 1 / 3 + 2 2 / 3 ) 3 = 2 + 2 2 + 3 ⋅ 2 1 / 3 ⋅ 2 2 / 3 ( 2 1 / 3 + 2 2 / 3 ⏟ x − 2 ) \begin{aligned}\underbrace{(x-2)^3}_{x^3-6x^2+12x-8}=&(2^{1/3}+2^{2/3})^3\\=&\small{2+2^2+3\cdot 2^{1/3}\cdot 2^{2/3}(\underbrace{2^{1/3}+2^{2/3}}_{x-2})}\end{aligned}

Rearranging we get:-

x 3 − 6 x 2 + 6 x = 2 x^3-6x^2+6x=\large\boxed{2}

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Yeah :) :)

Chirayu Bhardwaj - 4 years, 11 months ago

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:-)................

Rishabh Jain - 4 years, 11 months ago
Razing Thunder
Jul 3, 2020 
1
2
 
x=(2)+(2**(1/3)) + (2**(2/3))
print(x**3 - 6*x**2 + 6*x)

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