Answer:
y=10
IHJ is just a smaller size, the angles are the same.
How many times can 50 go in 70?
![\dfrac{\sqrt[3]{x+h}-\sqrt[3]x}h\times\dfrac{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}=\dfrac{(\sqrt[3]{x+h})^3-(\sqrt[3]x)^3}\cdots](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7Bx%2Bh%7D-%5Csqrt%5B3%5Dx%7Dh%5Ctimes%5Cdfrac%7B%5Csqrt%5B3%5D%7B%28x%2Bh%29%5E2%7D%2B%5Csqrt%5B3%5D%7Bx%28x%2Bh%29%7D%2B%5Csqrt%5B3%5D%7Bx%5E2%7D%7D%7B%5Csqrt%5B3%5D%7B%28x%2Bh%29%5E2%7D%2B%5Csqrt%5B3%5D%7Bx%28x%2Bh%29%7D%2B%5Csqrt%5B3%5D%7Bx%5E2%7D%7D%3D%5Cdfrac%7B%28%5Csqrt%5B3%5D%7Bx%2Bh%7D%29%5E3-%28%5Csqrt%5B3%5Dx%29%5E3%7D%5Ccdots)

The

s then cancel, leaving you with the
![\cdots=\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}](https://tex.z-dn.net/?f=%5Ccdots%3D%5Csqrt%5B3%5D%7B%28x%2Bh%29%5E2%7D%2B%5Csqrt%5B3%5D%7Bx%28x%2Bh%29%7D%2B%5Csqrt%5B3%5D%7Bx%5E2%7D)
term.
If it's not clear what I did above, consider the substitution
![a=\sqrt[3]{x+h}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7Bx%2Bh%7D)
and
![b=\sqrt[3]x](https://tex.z-dn.net/?f=b%3D%5Csqrt%5B3%5Dx)
. Then

Is there a picture to go along with this? This question is very vague. If you give me more info I might be able to help.
Find the difference between each number:
-24/4 = -6
144/-24 = -6
The next number is multiplied by -6
You have the first 4 terms
5th term = -864 x -6 = 5184
6th term = 5184 x -6 = -31,104
7th term = -31,104 x -6 = 186,624
8th term = 186,624 x -6 = -1,119,744