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How do you find all real number solutions to $\sqrt[3]{6 y + 2} - 2 = 0$ $root3(6y+2)-2=0$ ?

1 Answer

Nov 16, 2016

$y = 6$ $y=6$

Explanation:

your equation $\Rightarrow \sqrt[3]{6 y + 2} - 2 = 0$ $\rArr root3(6y+2)-2=0$

add 2 to both sides
$\sqrt[3]{6 y + 2} = 2$ $root3(6y+2)=2$
cube both sides to get rid of cube-root
$6 y + 2 = 2^{3}$ $6y+2=2^3$
simplify the cube
$6 y + 2 = 8$ $6y+2=8$
subtract 2 from both sides
$6 y = 8 - 2$ $6y=8-2$
simplify
$6 y = 6$ $6y=6$
divide both sides by 6 to isolate $y$ $y$ on left side
$y = 6$ $y=6$

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