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# Help

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$$\frac{1}{r^3+7}-7=\frac{-r^3}{r^3+7}$$

Solve for all values of r

Jun 16, 2019

#1
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Solve for r:
1/(r^3 + 7) - 7 = -r^3/(r^3 + 7)

Multiply both sides by r^3 + 7:
1 - 7 (r^3 + 7) = -r^3

Expand out terms of the left hand side:
-7 r^3 - 48 = -r^3

Add r^3 + 48 to both sides:
-6 r^3 = 48

Divide both sides by -6:
r^3 = -8

Taking cube roots gives 2 (-1)^(1/3) times the third roots of unity:
r = -2      or      r = 2 (-1)^(1/3)      or      r = -2 (-1)^(2/3)

Jun 16, 2019
#2
+7711
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$$\dfrac{1}{r^3 + 7} - 7 = \dfrac{-r^3}{r^3 + 7}\\ 1 - 7(r^3 + 7) = -r^3, r \neq \sqrt[3]{7}\\ 1 - 7r^3 - 49 = -r^3, r \neq \sqrt[3]{7}\\ 6r^3 = -48, r \neq \sqrt[3]{7}\\ r^3 = -8\\ r = -2, -2\omega, -2\omega^2\text{ where }\omega\text{ and }\omega^2\text{ denotes the cube roots of unity.}$$

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Jun 22, 2019