Question

Question 1
a^2 = 64 and b^3 = 64.

Solve a^2 = 64: a = ±√64 = ±8. Both 8 and −8 work because 8^2 = 64 and (−8)^2 = 64.
Solve b^3 = 64: b = ∛64 = 4. Only 4 works because 4^3 = 64 while (−4)^3 = −64.

Why they differ: an even exponent (2) loses sign information, so a can be positive or negative; an odd exponent (3) preserves sign, so b has a single real value.

Question 2
Solve 9(3x − 16) + 15 = 6x − 24.

Step 1 — distribute: 9·3x = 27x and 9·(−16) = −144,
so 27x − 144 + 15 = 6x − 24.

Step 2 — combine like terms on left: −144 + 15 = −129,
so 27x − 129 = 6x − 24.

Step 3 — subtract 6x from both sides: 21x − 129 = −24.

Step 4 — add 129 to both sides: 21x = 105.

Step 5 — divide by 21: x = 105/21 = 5.

Check: left = 9(3·5 −16)+15 = 9(15−16)+15 = 9(−1)+15 = 6; right = 6·5 −24 = 6. So x = 5.
part of question 2 says what value of x makes the equation true show the check to prove your solution is correct

Answer 1

Your answers are correct. Summary:

Question 1
- a^2 = 64 → a = ±√64 = ±8 (both 8 and −8 work because (±8)^2 = 64).
- b^3 = 64 → b = ∛64 = 4 (the real cube root is 4 because 4^3 = 64; (−4)^3 = −64, so −4 does not work).
Why they differ: even exponents lose sign information (both + and − give the same square), while odd exponents preserve sign (cube of a negative is negative), so a has two real solutions and b has one real solution.

Question 2
Solve 9(3x − 16) + 15 = 6x − 24:
- Distribute: 27x − 144 + 15 = 6x − 24
- Combine like terms: 27x − 129 = 6x − 24
- Subtract 6x: 21x − 129 = −24
- Add 129: 21x = 105
- Divide by 21: x = 5

Check:
Left side = 9(3·5 − 16) + 15 = 9(15 − 16) + 15 = 9(−1) + 15 = −9 + 15 = 6
Right side = 6·5 − 24 = 30 − 24 = 6
Since left = right = 6, x = 5 is correct.