What is the value of ...

Algebra Level 3

If x = ( 2 a + 4 a 2 − b 2 ) 1 3 + ( 2 a − 4 a 2 − b 2 ) 1 3 x=\left(2a+\sqrt{4a^2-b^2}\right)^\frac 13 + \left(2a-\sqrt{4a^2-b^2}\right)^\frac 13 , then

x 3 − 3 b 2 3 x + 9 a = ? \large x^3 - 3b^\frac 23 x + 9a =\ ?

None of the rest 14 a 14a 13 a 13a 12 a 12a

Akshit Sinha by Akshit Sinha , 18, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Chew-Seong Cheong
Mar 14, 2018

x = ( 2 a + 4 a 2 − b 2 ) 1 3 + ( 2 a − 4 a 2 − b 2 ) 1 3 Cubing both sides x 3 = 2 a + 4 a 2 − b 2 + 3 ( 2 a + 4 a 2 − b 2 ) 2 3 ( 2 a − 4 a 2 − b 2 ) 1 3 + 3 ( 2 a + 4 a 2 − b 2 ) 1 3 ( 2 a − 4 a 2 − b 2 ) 2 3 + 2 a − 4 a 2 − b 2 = 4 a + 3 ( 2 a + 4 a 2 − b 2 ) 1 3 ( 2 a − 4 a 2 − b 2 ) 1 3 ( ( 2 a + 4 a 2 − b 2 ) 1 3 + ( 2 a − 4 a 2 − b 2 ) 1 3 ) = 4 a + 3 ( 4 a 2 − 4 a 2 + b 2 ) 1 3 x = 4 a + 3 b 2 3 x \begin{aligned} x & = \left(2a+\sqrt{4a^2-b^2}\right)^\frac 13 + \left(2a-\sqrt{4a^2-b^2}\right)^\frac 13 \quad \quad \small \color{#3D99F6} \text{Cubing both sides} \\ x^3 & = 2a+\sqrt{4a^2-b^2} + 3 \left(2a+\sqrt{4a^2-b^2}\right)^\frac 23 \left(2a-\sqrt{4a^2-b^2}\right)^\frac 13 + 3 \left(2a+\sqrt{4a^2-b^2}\right)^\frac 13 \left(2a-\sqrt{4a^2-b^2}\right)^\frac 23 + 2a-\sqrt{4a^2-b^2} \\ & = 4a + 3 \left(2a+\sqrt{4a^2-b^2}\right)^\frac 13 \left(2a-\sqrt{4a^2-b^2}\right)^\frac 13 \color{#3D99F6}\left(\left(2a+\sqrt{4a^2-b^2}\right)^\frac 13 + \left(2a-\sqrt{4a^2-b^2}\right)^\frac 13 \right) \\ & = 4a + 3 \left(4a^2 - 4a^2+ b^2\right)^\frac 13 \color{#3D99F6}x \\ & = 4a + 3b^\frac 23 \color{#3D99F6}x \end{aligned}

⟹ x 3 − 3 b 2 3 x = 4 a x 3 − 3 b 2 3 x + 9 a = 13 a \begin{aligned} \implies x^3 - 3b^\frac 23 x & = 4a \\ x^3 - 3b^\frac 23 x + 9a & = \boxed{13a} \end{aligned}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...