#1**+1 **

\(\frac{1}{64a^3 + 7} - 7 = 0\)

Add 7 to both sides: \(\frac{1}{64a^3 + 7} =7\)

Multiply both sides by 64a^3 + 7 to get rid of denominator: \(1 = 7(64a^3+7)\); \(1 = 448a^3+49\); \(-48 =448a^3\); \(-{3\over28}=a^3\)

Cube rooting both sides, you get \(a = ({-{3\over28}})^{1\over3}\), or noted as the cube root of -3/28.

proyaop Feb 20, 2024

#1**+1 **

Best Answer

\(\frac{1}{64a^3 + 7} - 7 = 0\)

Add 7 to both sides: \(\frac{1}{64a^3 + 7} =7\)

Multiply both sides by 64a^3 + 7 to get rid of denominator: \(1 = 7(64a^3+7)\); \(1 = 448a^3+49\); \(-48 =448a^3\); \(-{3\over28}=a^3\)

Cube rooting both sides, you get \(a = ({-{3\over28}})^{1\over3}\), or noted as the cube root of -3/28.

proyaop Feb 20, 2024