Question

Using properties of proportion find x:y if x^3+12x/6x2+8=y^3+27y/9y2+27

Answer 1

Step-by-step explanation:

• Given that

        $\mathbf{\frac{x^{3}+12x}{6x^{2}+8}=\frac{y^{3}+27y}{9y^{2}+27}}$

        Using componendo and dividendo rules in above equation

        $\mathbf{\frac{x^{3}+12x+6x^{2}+8}{x^{3}+12x-6x^{2}-8}=\frac{y^{3}+27y+9y^{2}+27}{y^{3}+27y-9y^{2}-27}}$

        $\mathbf{\frac{x^{3}+6x^{2}+12x+8}{x^{3}-6x^{2}+12x-8}=\frac{y^{3}+9y^{2}+27y+27}{y^{3}-9y^{2}+27y-27}}$       ...1)

• From identity

        $\mathbf{(A+B)^{3}=A^{3}+3A^{2}B+3AB^{2}+B^{3}}$

        and

        $\mathbf{(A-B)^{3}=A^{3}-3A^{2}B+3AB^{2}-B^{3}}$

• Using identity in equation 1), we get

        $\mathbf{\frac{(x+2)^{3}}{(x-2)^{3}}=\frac{(y+3)^{3}}{(y-3)^{3}}}$        ...2)

• On taking cube root on both side of equation 2), we get

        $\mathbf{\frac{(x+2)}{(x-2)}=\frac{(y+3)}{(y-3)}}$

        On doing cross- multiplication

       $\mathbf{(x+2)(y-3)=(x-2)(y+3)}$

       $\mathbf{xy+2y-3x-6=xy-2y+3x-6}$

       $\mathbf{2y-3x=-2y+3x}$

       $\mathbf{2y+2y=3x+3x}$

       $\mathbf{4y=6x}$

       So

       $\mathbf{\frac{x}{y}=\frac{4}{6}=\frac{2}{3}=}$ This is answer.