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Find all real numbers $a$ that satisfy \[\frac{1}{a^3 + 7} -7 = \frac{-a^3}{a^3 + 7}.\]

Sagot :

The real number of the given expression is a=-2.

Given that the expression is [tex]\frac{1}{a^3+7}-7=\frac{-a^3}{a^3+7}[/tex]

Real numbers is known as the union of both rational and irrational numbers. Real numbers can be both positive or negative and are denoted by the symbol “R”.

To compute the value of a.

In the given expression multiply both sides with (a³+7) as

[tex]\frac{(a^3+7)}{a^3+7}-7=\frac{-a^3(a^3+7)}{a^3+7}[/tex]

Simplify the expression as,

1-7(a³+7)=-a³

Expand out terms of the left hand side

1-7a³-49=-a³

-7a³-48=-a³

Taking out minus common from both sides and cancel them

-(7a³+48)=-(a³)

7a³+48=a³

Add -a³-48 on both sides

7a³+48-a³-48=a³-a³-48

6a³=-48

Divide both sides with 6

(6÷6)a³=(-48÷6)

a³=-8

write -8 in the expansion form which is (-2)×(-2)×(-2)=-8

a³=(-2)³

From above it says that

a=-2

Hence, real number that satisfy [tex]\frac{1}{a^3+7}-7=\frac{-a^3}{a^3+7}[/tex] is a=-2.

Learn about real numbers from here brainly.com/question/10180279

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