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Solve. ^3√(3x+4)+6=7 (Not sure how to type a cube root, but that's what it is. Help.)
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thats readable. cbrt(3x+4) is workable as well; or 3rt(arg)
\[\sqrt[3]{3x+4}+6=7\]\[\sqrt[3]{3x+4}=7-6=1\]\[3x+4=1^{3}=1\]\[3x=1-4=-3\]\[x=-3/3=-1\]
subtract 6 from each side; cube it all to undo the cbrt and finish off to get to the x
then dbl chk lol
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Thanks to you both! :)
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