Answer:
Im doing it right now one second
Step-by-step explanation:
Start by decomposing the number inside the root into primes
Then group the terms into cubes if possible
rewrite the root
![\sqrt[3]{80}=\sqrt[3]{10\cdot2^3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B80%7D%3D%5Csqrt%5B3%5D%7B10%5Ccdot2%5E3%7D)
then cancel the terms that are cubes and bring them out of the root
![\sqrt[3]{80}=2\sqrt[3]{10}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B80%7D%3D2%5Csqrt%5B3%5D%7B10%7D)
Yes because you only need a 90 degree to become a right
Answer:
C. Kalena made a mistake in Step 3. The justification should state: -x²
+ x²
Step-by-step explanation:
Given the function x(x - 1)(x + 1) = x3 - X
To justify kelena proof
We will need to show if the two equations are equal.
Starting from the RHS with function x³-x
First we will factor out the common factor which is 'x' to have;
x(x²-1)
Factorising x²-1 using the difference of two square will give;
x(x+1)(x-1)
Note that for two real number a and b, the expansion of a²-b² using difference vof two square will give;
a²-b² = (a+b)(a-b) hence;
Factorising x²-1 using the difference of two square will give;
x(x+1)(x-1)
Factorising x(x+1) gives x²+x, therefore
x(x+1)(x-1) = (x²+x)(x-1)
(x²+x)(x-1) = x³-x²+x²-x
The function x³-x²+x²-x gotten shows that kelena made a mistake in step 3, the justification should be -x²+x² not -x-x²
