Question

How do you simplify `root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3))` ?

Answer 1

`root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3)) = 6`

Explanation:

Let:

`x = root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3))`

Then:

`x^3 = (135+78sqrt(3))+(135-78sqrt(3))+3(root(3)(135+78sqrt(3))root(3)(135-78sqrt(3)))(root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3)))`

`color(white)(x^3) = 270+3(root(3)(135^2-78^2(3)))x`

`color(white)(x^3) = 270+3(root(3)(18225-18252))x`

`color(white)(x^3) = 270+3(root(3)(-27))x`

`color(white)(x^3) = 270-9x`

So:

`x^3+9x-270 = 0`

By the rational roots theorem, any rational roots of this cubic are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-270` and `q` a divisor of the coefficient `1` of the leading term.

So the first few possible rational zeros are:

`+-1, +-2, +-3, +-5, +-6, +-9, +-10,...`

By Descartes' Rule of Signs, we can tell that the cubic has exactly one positive real root and no negative real zeros, so try the possible positive rational zeros.

We find:

`(color(blue)(6))^3+9(color(blue)(6))-270 = 216+54-270 = 0`

So `x=6` is a root and `(x-6)` is a factor:

`x^3+9x-270 = (x-6)(x^2+6x+45)`

The remaining quadratic has no real zeros (as we can see by looking at its discriminant).

The non-real zeros are actually:

`omega root(3)(135+78sqrt(3))+omega^2 root(3)(135-78sqrt(3))`

and:

`omega^2 root(3)(135+78sqrt(3))+omega root(3)(135-78sqrt(3))`

where `omega = -1/2+sqrt(3)/2i` is the primitive complex cube root of `1`.

So:

`root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3)) = 6`

Answer 2

`6`.

Explanation:

Let us note that,

`(x+sqrt3y)^3=x^3+3sqrt3y^3+3sqrt3xy(x+sqrt3y)`,

`=x^3+3sqrt3y^3+3sqrt3x^2y+9xy^2`.

`rArr (x+sqrt3y)^3=(x^3+9xy^2)+3sqrt3(x^2y+y^3), or,`

`root(3){(x^3+9xy^2)+sqrt3(3x^2y+3y^3)}=x+sqrt3y`.

So, from this we guess :

If `root(3)(135+78sqrt3)=x+sqrt3y`, then,

`(x^3+9xy^2)=135, and, (3x^2y+3y^3)=78, i.e.,`

`x^3+9xy^2=x(x^2+9y^2)=135......(1), and,`

`3y(x^2+y^2)=78, or, y(x^2+y^2)=26......(2)`.

`y(x^2+y^2)=26=1xx26, or, 2xx13`

By Trial and Error Method, `y=1, x=5;" not suitable for "(1)`

`y=2, x=3" satisfy "(1)`.

Therefore, `root(3)(135+78sqrt3)=x+sqrt3y=3+2sqrt3`.

Consequently, `root(3)(135-78sqrt3)=3-2sqrt3`.

Finally, `root(3)(135+78sqrt3)+root(3)(135-78sqrt3)=6`,

as Respected George C. Sir has already derived!