### Video Transcript

Evaluate the cube root of negative
27 plus the cube root of negative 0.125 plus the square root of five and one
sixteenth.

In this question, we are asked to
evaluate an expression involving three terms. The first term is the cube root of
an integer. The second term is the cube root of
a decimal. And the third term is the square
root of a mixed number.

To evaluate this expression, we can
start by recalling that it is often easiest to evaluate the roots of noninteger
values when they are written as fractions. So we will start by converting
negative 0.125 and five and one sixteenth into fractions. If we do this, then we can
calculate that negative 0.125 is negative one-eighth and five and one sixteenth is
81 over 16. This allows us to rewrite the
expression as the cube root of negative 27 plus the cube root of negative one-eighth
plus the square root of 81 over 16.

We now need to evaluate this
expression, and we will do this by evaluating each term separately. Let’s start with the first term,
the cube root of negative 27. We can evaluate this term by noting
that negative 27 is a perfect cube. In particular, negative 27 is equal
to negative three cubed. We can then evaluate this term by
recalling that for any real number 𝑎, we have that the cube root of 𝑎 cubed is
equal to 𝑎. So, the cube root of negative three
cubed is equal to negative three.

We can evaluate the second term by
using the laws of exponents. We can recall that for any real
numbers 𝑎 and 𝑏, with 𝑏 not equal to zero, we have that the cube root of 𝑎 over
𝑏 is equal to the cube root of 𝑎 over the cube root of 𝑏. Since negative one-eighth is equal
to negative one over eight, we can use this result with 𝑎 equal to negative one and
𝑏 equal to eight to rewrite the second term as the cube root of negative one over
the cube root of eight. We can then evaluate this term by
noting that negative one is equal to negative one cubed and eight is equal to two
cubed. Therefore, the numerator is equal
to negative one and the denominator is equal to two. So this term evaluates to give
negative one-half.

We can apply a similar process to
the third term. We know that the square root of 𝑎
over 𝑏 is equal to the square root of 𝑎 over the square root of 𝑏, though we do
need to be careful since we need 𝑎 to be nonnegative and 𝑏 to be positive. We can apply this result with 𝑎
equal to 81 and 𝑏 equal to 16 to get root 81 over root 16. Finally, we can evaluate the square
root of the numerator and denominator. We know that root 81 is nine and
root 16 is four. So we have negative three minus
one-half plus nine-quarters.

All that is left to do is evaluate
the expression. We rewrite the terms to have the
same denominator and evaluate to obtain negative five-quarters.