Must it be a cubic?

Algebra Level 3

x 3 − 12 x 2 + 36 x + 8 \large x^3 - 12x^2 + 36 x + 8

If x = 4 + 4 1 / 3 + 4 2 / 3 x = 4 + 4^{1/3} + 4^{2/3} , find the value of the expression above.


The answer is 44.

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Raven Herd by Raven Herd , 21, India

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2 solutions

x − 4 = 4 3 + 16 3 x-4=\sqrt[3]{4}+\sqrt[3]{16}

Now, cube both sides:

x 3 − 12 x 2 + 48 x − 64 = 4 + 3 4 3 16 3 ( 4 3 + 16 3 ) + 16 x 3 − 12 x 2 + 48 x − 64 = 20 + 3 64 3 ( x − 4 ) x 3 − 12 x 2 + 48 x − 64 = 20 + 12 x − 48 x 3 − 12 x 2 + 36 x − 36 = 0 x^3-12x^2+48x-64=4+3\sqrt[3]{4}\sqrt[3]{16}(\sqrt[3]{4}+\sqrt[3]{16})+16 \\ x^3-12x^2+48x-64=20+3\sqrt[3]{64}(x-4) \\ x^3-12x^2+48x-64=20+12x-48 \\ x^3-12x^2+36x-36=0

Finally, add 44 44 to both sides:

x 3 − 12 x 2 + 36 x + 8 = 44 x^3-12x^2+36x+8=\boxed{44}

14 Helpful 0 Interesting 1 Brilliant 0 Confused

There's a comparatively simpler way. We have,

x = 4 1 / 3 + 4 2 / 3 + 4 = 4 1 / 3 ( 1 + 4 1 / 3 + 4 2 / 3 ) = 4 1 / 3 ( 1 + x − 4 ) ⟹ x 3 = ( 4 1 / 3 ) 3 ⋅ ( x − 3 ) 3 = 4 ( x − 3 ) 3 x=4^{1/3}+4^{2/3}+4=4^{1/3}\left(1+4^{1/3}+4^{2/3}\right)=4^{1/3}(1+x-4)\\\implies x^3=(4^{1/3})^3\cdot (x-3)^3=4(x-3)^3

Expanding the RHS using binomial theorem and taking all terms to one side of equation, we immediately get,

x 3 − 12 x 2 + 36 x − 36 = 0 ⟹ x 3 − 12 x 2 + 36 x + 8 = 36 + 8 = 44 x^3-12x^2+36x-36=0\implies x^3-12x^2+36x+8=36+8=\boxed{44}

Prasun Biswas - 5 years, 12 months ago

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Yeah, you always simplify everything.

Alan Enrique Ontiveros Salazar - 5 years, 12 months ago

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Haha, I just try to. :P

Prasun Biswas - 5 years, 12 months ago

nice solution.

Raven Herd - 5 years, 11 months ago

thank you for the solution but where did you get the (x-3) from thank you

kool guy - 5 years, 1 month ago

@Prasun Biswas RHS means the Right angle - Hypotenuse - Side congruence of the condition of being congruent with the right triangle.

. . - 3 months ago
Chew-Seong Cheong
Jun 14, 2015

x 3 − 12 x 2 + 36 x + 8 = x 3 − 3 ( 4 ) x 2 + 3 ( 4 2 ) x + 4 3 − 12 x + 72 = ( x − 4 ) 3 − 12 ( x − 4 ) + 24 = ( 4 3 + 16 3 ) 3 − 12 ( 4 3 + 16 3 ) + 24 = 4 + 3 ( 4 3 ) 2 ( 16 3 ) + 3 ( 4 3 ) ( 16 3 ) 2 + 16 − 12 4 3 − 12 16 3 + 24 = 4 + 12 4 3 + 12 16 3 + 16 − 12 4 3 − 12 16 3 + 24 = 44 \begin{aligned} x^3-12x^2+36x+8 & = x^3 - 3(4)x^2+3(4^2)x + 4^3 - 12 x + 72 \\ & = (x-4)^3 - 12(x-4) + 24 \\ & = (\sqrt[3]{4} + \sqrt[3]{16})^3 - 12(\sqrt[3]{4} + \sqrt[3]{16}) + 24 \\ & = 4 + 3(\sqrt[3]{4})^2(\sqrt[3]{16}) + 3(\sqrt[3]{4})(\sqrt[3]{16})^2 + 16 \\ & \quad \quad - 12\sqrt[3]{4} - 12\sqrt[3]{16} + 24 \\ & = 4 + 12\sqrt[3]{4} + 12\sqrt[3]{16} + 16 - 12\sqrt[3]{4} - 12\sqrt[3]{16} + 24 \\ & = \boxed{44} \end{aligned}

9 Helpful 0 Interesting 0 Brilliant 0 Confused

( a + b ) 3 = a 3 + a 2 b + a b 2 + b 3 ( a + b ) ^ { 3 } = a ^ { 3 } + a ^ { 2 } b + ab ^ { 2 } + b ^ { 3 } , and ( a − b ) 3 = a 3 − a 2 b + a b 2 − b 3 ( a - b ) ^ { 3 } = a ^ { 3 } - a ^ { 2 } b + ab ^ { 2 } - b ^ { 3 } .

Then if ( x − 4 ) 3 ( x - 4 ) ^ { 3 } , then it must be x 3 − 3 ( 4 ) x 2 + 3 ( 4 2 ) x − 4 3 − 12 x + 72 x ^ { 3 } - 3 ( 4 ) x ^ { 2 } + 3 ( 4 ^ { 2 } ) x \boxed { - 4 ^ { 3 } } - 12x + 72 .

. . - 3 months ago

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