Can it be an integer?

Algebra Level 3

u = 4 + 44 5 5 3 , v = 4 44 5 5 3 u=\sqrt[3]{4+\frac{44}{5\sqrt{5}}}, \,\,\,\,\, v=\sqrt[3]{4-\frac{44}{5\sqrt{5}}}

Suppose the values of u u and v v are as stated. Find the value of 5 ( u + v ) 3 12 ( u + v ) . 5\left(u+v\right)^3-12\left(u+v\right).


The answer is 40.

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2 solutions

Chew-Seong Cheong
Oct 30, 2017

Similar solution with @Chan Lye Lee

( u + v ) 3 = u 3 + v 3 + 3 u v ( u + v ) = 4 + 44 5 5 + 4 44 5 5 + 3 4 + 44 5 5 3 4 44 5 5 3 ( u + v ) = 8 + 3 ( 4 + 44 5 5 ) ( 4 44 5 5 ) 3 ( u + v ) = 8 + 3 4 2 4 4 2 ( 5 5 ) 2 3 ( u + v ) = 8 + 3 64 125 3 ( u + v ) = 8 + 12 5 ( u + v ) Multiply both sides by 5 5 ( u + v ) 3 = 40 + 12 ( u + v ) \begin{aligned} (u+v)^3 & = u^3+v^3+3uv(u+v) \\ & = 4 + \frac {44}{5\sqrt 5} + 4 - \frac {44}{5\sqrt 5} + 3 \sqrt[3]{4 + \frac {44}{5\sqrt 5}} \sqrt[3]{4 - \frac {44}{5\sqrt 5}} (u+v) \\ & = 8 + 3 \sqrt[3]{\left(4 + \frac {44}{5\sqrt 5}\right) \left(4 - \frac {44}{5\sqrt 5}\right)} (u+v) \\ & = 8 + 3 \sqrt[3]{4^2 - \frac {44^2}{(5\sqrt 5)^2}} (u+v) \\ & = 8 + 3 \sqrt[3]{\frac {64}{125}} (u+v) \\ & = 8 + \frac {12}5 (u+v) & \small \color{#3D99F6} \text{Multiply both sides by 5} \\ 5(u+v)^3 & = 40 + 12(u+v) \end{aligned}

5 ( u + v ) 3 12 ( u + v ) = 40 \implies 5(u+v)^3 - 12(u+v) = \boxed{40}

Chan Lye Lee
Oct 30, 2017

We first note that u v = 4 + 44 5 5 3 × 4 44 5 5 3 = 4 2 4 4 2 5 2 × 5 3 = 4 2 125 ( 125 1 1 2 ) 3 = 4 5 uv=\sqrt[3]{4+\frac{44}{5\sqrt{5}}} \times \sqrt[3]{4-\frac{44}{5\sqrt{5}}} =\sqrt[3]{4^2-\frac{44^2}{5^2 \times 5}}= \sqrt[3]{ \frac{4^2}{125} \left(125-11^2\right)} = \frac{4}{5} .

Next, u 3 = 4 + 44 5 5 u^3=4+\frac{44}{5\sqrt{5}} and v 3 = 4 44 5 5 v^3=4-\frac{44}{5\sqrt{5}} , which imply that u 3 + v 3 = 8 u^3+v^3=8 .

Finally, 5 ( u + v ) 3 12 ( u + v ) = 5 ( u 3 + 3 u v ( u + v ) + v 3 ) 12 ( u + v ) = 5 ( 8 + 3 × 4 5 ( u + v ) ) 12 ( u + v ) = 40. 5\left(u+v\right)^3-12\left(u+v\right) = 5\left(u^3+3uv(u+v)+v^3\right)-12\left(u+v\right) = 5\left(8+3\times \frac{4}{5}(u+v)\right) -12\left(u+v\right)=40.

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