If (x^{3}-y^{3}):(x^{2}+xy+y^{2})=5:1 and (x^ | vedclass

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(B) Given $\frac{x^{3}-y^{3}}{x^{2}+xy+y^{2}}=\frac{5}{1}$.Using the algebraic identity $x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})$,we get $\frac{(x-y)(x^{2}+

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$4:1$

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(B) Given $\frac{x^{3}-y^{3}}{x^{2}+xy+y^{2}}=\frac{5}{1}$.Using the algebraic identity $x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})$,we get $\frac{(x-y)(x^{2}+xy+y^{2})}{x^{2}+xy+y^{2}}=5$,which simplifies to $x-y=5$ (Equation $1$).Given $\frac{x^{2}-y^{2}}{x-y}=\frac{7}{1}$.Using the identity $x^{2}-y^{2}=(x+y)(x-y)$,we get $\frac{(x+y)(x-y)}{x-y}=7$,which simplifies to $x+y=7$ (Equation $2$).Adding Equation $1$ and Equation $2$: $(x-y)+(x+y)=5+7 \Rightarrow 2x=12 \Rightarrow x=6$.Substituting $x=6$ into Equation $2$: $6+y=7 \Rightarrow y=1$.Therefore,the ratio $\frac{2x}{3y} = \frac{2 	imes 6}{3 	imes 1} = \frac{12}{3} = \frac{4}{1}$.

If $(x^{3}-y^{3}):(x^{2}+xy+y^{2})=5:1$ and $(x^{2}-y^{2}):(x-y)=7:1$,then the ratio $2x:3y$ equals

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