Simplify Surds?

Question

Can u explain how to simplify without using decimals?
(sqrt 5 - sqrt 3)^3

Answer 1

One way to do this is, instead of actually cubing this, you can separate it into the following:

[sqrt(5) - sqrt(3)]^2 * [sqrt(5) - sqrt(3)]

Squaring a binomial is much simpler than cubing it. Now, we solve this. To make things simpler, I'll write out the binomial being squared, and then use FOIL so it can visually be seen what I'm doing.

[sqrt(5) - sqrt(3)][sqrt(5) - sqrt(3)] * [sqrt(5) - sqrt(3)]

Remember that any square root times itself eliminates the square root symbol.

[5 - sqrt(3)sqrt(5) - sqrt(3)sqrt(5) + 3] * [sqrt(5) - sqrt(3)]

Now, we can group together like-terms.

[8 - 2sqrt(3)sqrt(5)] [sqrt(5) - sqrt(3)]

And we use FOIL again.

8sqrt(5) - 8sqrt(3) - 2sqrt(3)[5] + 2(3)sqrt(5)
8sqrt(5) - 8sqrt(3) - 10sqrt(3) + 6sqrt(5)

Grouping together like-terms again,

14sqrt(5) - 18sqrt(3)

Which is now in its simplest form.

It is sometimes better to work with squares rather than cubes, especially when simplifying SQUARE roots. As a reminder, I separated something of the form
(x - y)^3
into
(x - y)^2 * (x - y)^1

because it's exactly the same.

Answer 2

Be careful when doing things like this,
because sqrt (a + b) is not equal to sqrt a + sqrt b. And (a + b)^n is not equal to a^n + b^n. These are common mistakes.
For a question like this, I would express all surds as fractional exponents, then use the binomial theorem. For small exponents like 3, you could get the coefficioents from Pascal's Triangle rather than having to bother with the combinations part of the binomial theorem. So your coefficients for the kth term would be C(n,(k-1)), with k = 3, that is, 1, 3, 3, 1
Now your terms look like this: a^(k)b^(n-(k-1))
Since a = sqrt 5 or 5^(1/2) and b = sqrt 3 or 3(1/2) you have for the first term
1x(5^(1/2))^(3) x (3^(1/2))^0)
Since raising a power to another power multiplies the exponents, and anything to the zero is 1, this first term simplifies to
5^(3/2) or 5 sqrt 5
Your next term is more interesting. It will be
3x(5^(1/2)^2)x(3^(1/2)^1) which is
3 x 5 x (3^(1/2) or 15 sqrt 3
You would go on to the next two terms this way, writing out all four terms. Finally, you would simplify further by gathering and combining like terms.

Answer 3

Expand the brackets. Note that, for example, (sqrt 5)^3 = 5^(3/2) = 5*sqrt(5).

Answer 4

Let x = sqrt(5)
and y = sqrt(3)

(x-y)^3 = x^3-3x^2y+3xy^2-y^3
So
(sqrt(5) - sqrt(3))^3 =
(sqrt(5))^3 -3(sqrt(5))^2*sqrt(3)
+3*sqrt(5)*(sqrt(3))^2
-(sqrt(5))^3=

5sqrt(5)-15sqrt(3)
+9sqrt(5)-3sqrt(3)=

(5+9)sqrt(5)-(15+3)sqrt(3) =

14*sqrt(5) - 18*sqrt(3)

Answer 5

wwefan76's answer is wrong. That isn't the proper way to expand the cube. barks answer is right, if you follow the proper rules to expand a cube, you will find what he says will give hints as to simplifying it.

But be careful about expanding the cube as it isn't as simple as wwefan says.

Answer 6

Let x = sqrt5 and y = sqrt3
(x-y)^3 = x^3 - 3(x^2)y +3x(y^2) - y^3
Remember x^2 = 5 and y^2 = 3
So 5x - 3*5y + 3*3x - 3y =
5x - 15y + 9x - 3y =
14x - 18y =
14*SQRT5 - 18*SQRT3

Answer 7

You need to cube both of them:
(√5)^3-(√3)^3
(√125)-(√27)
5√5-3√3
You can do the rest by yourself!
I hope this helps!