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Question Number 177929 by Spillover last updated on 11/Oct/22

Find LCM   3y+12,y^2 −16 and y^4 −64y

$$\mathrm{Find}\:\mathrm{LCM}\: \\ $$$$\mathrm{3y}+\mathrm{12},\mathrm{y}^{\mathrm{2}} −\mathrm{16}\:\mathrm{and}\:\mathrm{y}^{\mathrm{4}} −\mathrm{64y} \\ $$

Answered by Ar Brandon last updated on 11/Oct/22

3y+12, y^2 −16, y^4 −64y  ⇔3(y+4), (y−4)(y+4), y(y−4)(y^2 +4y+16)   determinant (((y+4),(3(y+4)),((y−4)(y+4)),(y(y−4)(y^2 +4y+16))),((y−4),3,(y−4),(y(y−4)(y^2 +4y+16))),((y^2 +4y+16),3,1,(y(y^2 +4y+16))),(y,3,1,y),(3,3,1,1),(,1,1,1))  L.C.M=3y(y−4)(y+4)(y^2 +4y+16)                 =3(y+4)(y^4 −64y)

$$\mathrm{3y}+\mathrm{12},\:\mathrm{y}^{\mathrm{2}} −\mathrm{16},\:\mathrm{y}^{\mathrm{4}} −\mathrm{64y} \\ $$$$\Leftrightarrow\mathrm{3}\left(\mathrm{y}+\mathrm{4}\right),\:\left(\mathrm{y}−\mathrm{4}\right)\left(\mathrm{y}+\mathrm{4}\right),\:\mathrm{y}\left(\mathrm{y}−\mathrm{4}\right)\left(\mathrm{y}^{\mathrm{2}} +\mathrm{4y}+\mathrm{16}\right) \\ $$$$\begin{array}{|c|c|c|c|c|c|}{\mathrm{y}+\mathrm{4}}&\hline{\mathrm{3}\left(\mathrm{y}+\mathrm{4}\right)}&\hline{\left(\mathrm{y}−\mathrm{4}\right)\left(\mathrm{y}+\mathrm{4}\right)}&\hline{\mathrm{y}\left(\mathrm{y}−\mathrm{4}\right)\left(\mathrm{y}^{\mathrm{2}} +\mathrm{4y}+\mathrm{16}\right)}\\{\mathrm{y}−\mathrm{4}}&\hline{\mathrm{3}}&\hline{\mathrm{y}−\mathrm{4}}&\hline{\mathrm{y}\left(\mathrm{y}−\mathrm{4}\right)\left(\mathrm{y}^{\mathrm{2}} +\mathrm{4y}+\mathrm{16}\right)}\\{\mathrm{y}^{\mathrm{2}} +\mathrm{4y}+\mathrm{16}}&\hline{\mathrm{3}}&\hline{\mathrm{1}}&\hline{\mathrm{y}\left(\mathrm{y}^{\mathrm{2}} +\mathrm{4y}+\mathrm{16}\right)}\\{\mathrm{y}}&\hline{\mathrm{3}}&\hline{\mathrm{1}}&\hline{\mathrm{y}}\\{\mathrm{3}}&\hline{\mathrm{3}}&\hline{\mathrm{1}}&\hline{\mathrm{1}}\\{}&\hline{\mathrm{1}}&\hline{\mathrm{1}}&\hline{\mathrm{1}}\\\hline\end{array} \\ $$$${L}.{C}.{M}=\mathrm{3y}\left(\mathrm{y}−\mathrm{4}\right)\left(\mathrm{y}+\mathrm{4}\right)\left(\mathrm{y}^{\mathrm{2}} +\mathrm{4y}+\mathrm{16}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3}\left(\mathrm{y}+\mathrm{4}\right)\left(\mathrm{y}^{\mathrm{4}} −\mathrm{64y}\right) \\ $$

Commented by Spillover last updated on 11/Oct/22

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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