x^1/3; x^10/3 + x^1/3
What are we doing with it?
the ; is not an arithmetic operator
Sorry, I forgot to indicate.
Factor the common factor from the expression:
x^1/3; x^10/3 + x^1/3
x^10/3 + x^1/3
= x^(1/3) (x^3 + 1)
so the common factor is x^(1/3)
To simplify the expression x^(1/3), you can rewrite it as the cube root of x.
To simplify the expression x^(10/3) + x^(1/3), you can use the property of exponents which states that x^(a+b) = x^a * x^b. Applying this property, you have x^(10/3) + x^(1/3) = x^(9/3 + 1/3) + x^(1/3) = x^(10/3) * x^(1/3) + x^(1/3) = x^((10/3) + (1/3)) + x^(1/3) = x^(11/3) + x^(1/3).
Now, both x^(11/3) and x^(1/3) are in the form x^(a/b), where a and b are integers. To add or subtract terms in this form, the bases (here, x) need to be the same. Therefore, you can rewrite the expression as:
x^(11/3) + x^(1/3) = (x^11)^(1/3) + x^(1/3)
Now, you have a common base of x and can add the exponents.
Thus, the simplified expression is (x^11)^(1/3) + x^(1/3).