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

 Feb 20, 2024

Best Answer 

 #1
avatar+1622 
+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.

 Feb 20, 2024
 #1
avatar+1622 
+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

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