Question

If `a^2+2=3^(2/3)+3^(-2/3)` then show `3a^3+9a=8`? Thanks in Advance.

Answer 1

Actually `3a^3+9a = +-8`

Explanation:

Note that:

`(3^(1/3)-3^(-1/3))^2 = (3^(1/3))^2-2(3^(1/3))(3^(-1/3)+(3^(-1/3))^2`

`color(white)((3^(1/3)-3^(-1/3))^2) = 3^(2/3)-2+3^(-2/3)`

So given:

`a^2+2 = 3^(2/3)+3^(-2/3)`

Subtract `2` from both sides to get:

`a^2 = 3^(2/3)-2+3^(-2/3) = (3^(1/3)-3^(-1/3))^2`

So:

`a = +-(3^(1/3)-3^(-1/3))`

If `a = 3^(1/3)-3^(-1/3)` then:

`3a^3+9a = 3a(a^2+3)`

`color(white)(3a^3+9a) = 3(3^(1/3)-3^(-1/3))((3^(1/3)-3^(-1/3))^2+3)`

`color(white)(3a^3+9a) = 3(3^(1/3)-3^(-1/3))(3^(2/3)+1+3^(-2/3))`

`color(white)(3a^3+9a) = 3(3^(1/3)(3^(2/3)+1+3^(-2/3))-3^(-1/3)(3^(2/3)+1+3^(-2/3)))`

`color(white)(3a^3+9a) = 3((3+3^(1/3)+3^(-1/3))-(3^(1/3)+3^(-1/3)+1/3))`

`color(white)(3a^3+9a) = 3(3-1/3) = 9-1 = 8`

If we reverse the sign of `a` then `3a^3+9a = -8`

Answer 2

`3x^3+9a = +-8`

Explanation:

Here's another way of doing this that is slightly less painful:

Let `t = 3^(1/3)`

Then:

`a^2+2 = t^2+1/t^2`

Adding `1` to both sides, we have:

`a^2+3 = t^2+1+1/t^2`

Instead, subtracting `2` from both sides, we get:

`a^2 = t^2-2+1/t^2 = (t-1/t)^2`

So:

`a = +-(t-1/t)`

Then:

`3a^3+9a = 3a(a^2+3)`

`color(white)(3a^3+9a) = +-3(t-1/t)(t^2+1+1/t^2)`

`color(white)(3a^3+9a) = +-3(t(t^2+1+1/t^2)-1/t(t^2+1+1/t^2))`

`color(white)(3a^3+9a) = +-3((t^3+t+1/t)-(t+1/t+1/t^3))`

`color(white)(3a^3+9a) = +-3(t^3-1/t^3)`

`color(white)(3a^3+9a) = +-3(3-1/3)`

`color(white)(3a^3+9a) = +-(9-1)`

`color(white)(3a^3+9a) = +-8`