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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+1633 
+1

164a3+77=0

Add 7 to both sides: 164a3+7=7

Multiply both sides by 64a^3 + 7 to get rid of denominator: 1=7(64a3+7)1=448a3+4948=448a3328=a3

 

Cube rooting both sides, you get a=(328)13, or noted as the cube root of -3/28.

 Feb 20, 2024
 #1
avatar+1633 
+1
Best Answer

164a3+77=0

Add 7 to both sides: 164a3+7=7

Multiply both sides by 64a^3 + 7 to get rid of denominator: 1=7(64a3+7)1=448a3+4948=448a3328=a3

 

Cube rooting both sides, you get a=(328)13, or noted as the cube root of -3/28.

proyaop Feb 20, 2024

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