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\[y ^{2/3}+12=7y ^{1/3}\]
HI!!
the gimmick is to think of this as a quadratic equation in \(y^{\frac{1}{3}}\)
put say \(x=y^{\frac{1}{3}}\) then solve \[x^2+12=x\]
solve as normal for a quadratic equation \[x^2-x+12=0\] factor etc then once you have solved for \(x\) replace \(x\) by \(y^{\frac{1}{3}}\) and solve for \(y\)
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you got this or no?
\[\sqrt[3]{-3}\]
\[\sqrt[3]{4}\]
you got it backwards i think
\[y^{\frac{1}{3}}=-3\\ y=(-3)^3\]
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also for the four
got that?
-27, 64
yes i think so
awesome thanks again :)
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\[\color\magenta\heartsuit\]
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