Evaluate the brace containing: [(2/3) cubed divided by (4/3) cubed plus 7/8 times (1/4) to the zero] squared, divided by 7/8, minus 1/7 - all of that divided by 4/5, then minus 1/2.

Step-by-step solution

1. 1

   Work strictly from the innermost grouping outward. The nesting is parentheses inside square brackets inside braces, so the square bracket must be finished before it can be squared, and the braces before the final division.

2. 2

   Divide the two cubes. $\left(\tfrac23\right)^3 = \tfrac{8}{27}$ and $\left(\tfrac43\right)^3 = \tfrac{64}{27}$, so $\frac{8}{27}:\frac{64}{27} = \frac{8}{64} = \frac18.$ The common denominator $27$ cancels, which is why the cubes are worth expanding here rather than using an exponent rule (the bases differ).

3. 3

   Handle the zero exponent. Any nonzero base to the power $0$ is $1$, so $\tfrac78\left(\tfrac14\right)^0 = \tfrac78$. The square bracket is therefore $\frac18+\frac78 = 1.$

4. 4

   Square it and divide. $1^2 = 1$, and $1:\frac78 = \frac87.$ Dividing by a fraction means multiplying by its reciprocal.

5. 5

   Close the braces. $\frac87-\frac17 = \frac77 = 1.$

6. 6

   Finish the outer operations. $1:\frac45 = \frac54, \qquad \frac54-\frac12 = \frac54-\frac24 = \frac34.$ Note the final subtraction comes last - it is outside every bracket.