The answer is 32 8times 4
Consider the equation 
.
First, you can use the substitution 
, then 
and equation becomes 
. This equation is quadratic, so
.
Then you can factor this equation:
.
Use the made substitution again:
.
You have in each brackets the expression like 
that is equal to 
. Thus,
![x^3+5=(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25}) ,\\x^3+1=(x+1)(x^2-x+1)](https://tex.z-dn.net/?f=%20x%5E3%2B5%3D%28x%2B%5Csqrt%5B3%5D%7B5%7D%29%28x%5E2-%5Csqrt%5B3%5D%7B5%7Dx%2B%5Csqrt%5B3%5D%7B25%7D%29%20%2C%5C%5Cx%5E3%2B1%3D%28x%2B1%29%28x%5E2-x%2B1%29%20%20%20)
and the equation is
![(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25})(x+1)(x^2-x+1)=0](https://tex.z-dn.net/?f=%20%28x%2B%5Csqrt%5B3%5D%7B5%7D%29%28x%5E2-%5Csqrt%5B3%5D%7B5%7Dx%2B%5Csqrt%5B3%5D%7B25%7D%29%28x%2B1%29%28x%5E2-x%2B1%29%3D0%20%20%20)
.
Here ![x_1=-\sqrt[3]{5} , x_2=-1](https://tex.z-dn.net/?f=%20x_1%3D-%5Csqrt%5B3%5D%7B5%7D%20%2C%20x_2%3D-1%20)
and you can sheck whether quadratic trinomials have real roots:
1. ![D_1=(-\sqrt[3]{5}) ^2-4\cdot \sqrt[3]{25}=\sqrt[3]{25} -4\sqrt[3]{25} =-3\sqrt[3]{25}](https://tex.z-dn.net/?f=%20D_1%3D%28-%5Csqrt%5B3%5D%7B5%7D%29%20%5E2-4%5Ccdot%20%5Csqrt%5B3%5D%7B25%7D%3D%5Csqrt%5B3%5D%7B25%7D%20-4%5Csqrt%5B3%5D%7B25%7D%20%3D-3%5Csqrt%5B3%5D%7B25%7D%20%20%3C0%20)
.
2. 
.
This means that quadratic trinomials don't have real roots.
Answer:
If you need complex roots, then
![x_{3,4}=\dfrac{\sqrt[3]{5}\pm i\sqrt{3\sqrt[3]{25}}}{2} ,\\ \\x_{5,6}=\dfrac{1\pm i\sqrt{3}}{2}](https://tex.z-dn.net/?f=%20x_%7B3%2C4%7D%3D%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%5Cpm%20i%5Csqrt%7B3%5Csqrt%5B3%5D%7B25%7D%7D%7D%7B2%7D%20%20%20%2C%5C%5C%20%5C%5Cx_%7B5%2C6%7D%3D%5Cdfrac%7B1%5Cpm%20i%5Csqrt%7B3%7D%7D%7B2%7D%20%20%20%20)
.
19/2 = 19 divided by 2 = 9.5
Hope this helped.
The sides of a triangle must satisfy the triangle inequality, which states the sum of the lengths of any two sides is strictly greater than the length of the remaining side.
We really only have to check if the sum of the two smaller sides exceeds the largest side.
A. 5+6>7, ok
B. 6+6>10, ok
C. 7+7=14 Not ok, this is a degenerate triangle not a real triangle
D. 4+6>8 ok
Answer: C
