Equation

Question

0

65

1

+789

Find all real numbers a that satisfy \frac{1}{64a^3 + 7} - 7 = 0.

bingboy Feb 20, 2024

+1633

+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

proyaop · Answer 1 · 2024-02-20T04:14:32

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

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.