Start by decomposing the number inside the root into primes
Then group the terms into cubes if possible
![\begin{gathered} 80=2\cdot2\cdot2\cdot2\cdot5 \\ 80=2^3\cdot2\cdot5 \\ 80=10\cdot2^3 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%2080%3D2%5Ccdot2%5Ccdot2%5Ccdot2%5Ccdot5%20%5C%5C%2080%3D2%5E3%5Ccdot2%5Ccdot5%20%5C%5C%2080%3D10%5Ccdot2%5E3%20%5Cend%7Bgathered%7D)
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
how do i answer that I don't even know that sorry
Answer:
10
Step-by-step explanation:
Step 1: Simplify both sides of the equation.
3x−5=
1
2
x+2x
3x+−5=
1
2
x+2x
3x−5=(
1
2
x+2x)(Combine Like Terms)
3x−5=
5
2
x
3x−5=
5
2
x
Step 2: Subtract 5/2x from both sides.
3x−5−
5
2
x=
5
2
x−
5
2
x
1
2
x−5=0
Step 3: Add 5 to both sides.
1
2
x−5+5=0+5
1
2
x=5
Step 4: Multiply both sides by 2.
2*(
1
2
x)=(2)*(5)
x=10