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Question Number 202258 by MATHEMATICSAM last updated on 23/Dec/23

If (x/a) = (y/b) then show that ((x^3 + 3xy^2 )/(a^3 + 3ab^2 )) = (( y^3 + 3x^2 y)/(b^3 + 3a^2 b)) .

$$\mathrm{If}\:\frac{{x}}{{a}}\:=\:\frac{{y}}{{b}}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\frac{{x}^{\mathrm{3}} \:+\:\mathrm{3}{xy}^{\mathrm{2}} }{{a}^{\mathrm{3}} \:+\:\mathrm{3}{ab}^{\mathrm{2}} }\:=\:\frac{\:{y}^{\mathrm{3}} \:+\:\mathrm{3}{x}^{\mathrm{2}} {y}}{{b}^{\mathrm{3}} \:+\:\mathrm{3}{a}^{\mathrm{2}} {b}}\:. \\ $$

Answered by deleteduser1 last updated on 23/Dec/23

((x(x^2 +3y^2 )=x^3 (1+((3y^2 )/x^2 )))/(y^3 (((3x^2 )/y^2 )+1)))=(a^3 /b^3 )×(((3b^2 +a^2 )/a^2 )/((3a^2 +b^2 )/b^2 )) =((a(3b^2 +a^2 ))/(b(3a^2 +b^2 )))=((a^3 +3ab^2 )/(3a^2 b+b^3 )).

$$\frac{{x}\left({x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} \right)={x}^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{3}{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)}{{y}^{\mathrm{3}} \left(\frac{\mathrm{3}{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }+\mathrm{1}\right)}=\frac{{a}^{\mathrm{3}} }{{b}^{\mathrm{3}} }×\frac{\frac{\mathrm{3}{b}^{\mathrm{2}} +{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}{\frac{\mathrm{3}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} }} \\ $$$$=\frac{{a}\left(\mathrm{3}{b}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}{{b}\left(\mathrm{3}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}=\frac{{a}^{\mathrm{3}} +\mathrm{3}{ab}^{\mathrm{2}} }{\mathrm{3}{a}^{\mathrm{2}} {b}+{b}^{\mathrm{3}} }. \\ $$

Answered by Rasheed.Sindhi last updated on 23/Dec/23

(x/a) = (y/b)=k x=ak,y=bk lhs: (((ak)^3 +3(ak)(bk)^2 )/(3(ak)^2 (bk)+(bk)^3 ))=((a^3 k^3 +3ab^2 k^3 )/(3a^2 bk^3 +b^3 k^3 )) =((k^3 (a^3 +3ab^2 ))/(k^3 (3a^2 b + b^3 )))=((a^3 + 3ab^2 )/(3a^2 b + b^3 ))=rhs

$$\frac{{x}}{{a}}\:=\:\frac{{y}}{{b}}={k} \\ $$$${x}={ak},{y}={bk} \\ $$$${lhs}: \\ $$$$\frac{\left({ak}\right)^{\mathrm{3}} +\mathrm{3}\left({ak}\right)\left({bk}\right)^{\mathrm{2}} }{\mathrm{3}\left({ak}\right)^{\mathrm{2}} \left({bk}\right)+\left({bk}\right)^{\mathrm{3}} }=\frac{{a}^{\mathrm{3}} {k}^{\mathrm{3}} +\mathrm{3}{ab}^{\mathrm{2}} {k}^{\mathrm{3}} }{\mathrm{3}{a}^{\mathrm{2}} {bk}^{\mathrm{3}} +{b}^{\mathrm{3}} {k}^{\mathrm{3}} } \\ $$$$=\frac{{k}^{\mathrm{3}} \left({a}^{\mathrm{3}} +\mathrm{3}{ab}^{\mathrm{2}} \right)}{{k}^{\mathrm{3}} \left(\mathrm{3}{a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{3}} \right)}=\frac{{a}^{\mathrm{3}} \:+\:\mathrm{3}{ab}^{\mathrm{2}} }{\mathrm{3}{a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{3}} }={rhs} \\ $$

Answered by Frix last updated on 23/Dec/23

((x(x^2 +3y^2 ))/(a(a^2 +3b^2 )))=((y(y^2 +3x^2 ))/(b(b^2 +3a^2 ))) (x/a)=(y/b) ((x^2 +3y^2 )/(a^2 +3b^2 ))=((y^2 +3x^2 )/(b^2 +3a^2 )) ((b^2 +3a^2 )/(a^2 +3b^2 ))=((y^2 +3x^2 )/(x^2 +3y^2 )) y=((bx)/a) ((b^2 +3a^2 )/(a^2 +3b^2 ))=((((b^2 x^2 )/a^2 )+3x^2 )/(x^2 +((3b^2 x^2 )/a^2 ))) true

$$\frac{{x}\left({x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} \right)}{{a}\left({a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} \right)}=\frac{{y}\left({y}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{2}} \right)}{{b}\left({b}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} \right)} \\ $$$$\frac{{x}}{{a}}=\frac{{y}}{{b}} \\ $$$$\frac{{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} }{{a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} }=\frac{{y}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{2}} }{{b}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} } \\ $$$$\frac{{b}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} }=\frac{{y}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} } \\ $$$${y}=\frac{{bx}}{{a}} \\ $$$$\frac{{b}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} }=\frac{\frac{{b}^{\mathrm{2}} {x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\mathrm{3}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\frac{\mathrm{3}{b}^{\mathrm{2}} {x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\:\:\:\:\:\mathrm{true} \\ $$

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