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If-2x-a-4b-3-a-3-3a-and-2y-a-4b-3-a-3-3a-then-show-that-x-3-y-3-b-3-

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Question Number 202353 by MATHEMATICSAM last updated on 25/Dec/23
If 2x = a + (√((4b^3 − a^3 )/(3a))) and 2y = a − (√((4b^3 − a^3 )/(3a))) then show that x^3 + y^3 = b^3 .
$$\mathrm{If}\:\mathrm{2}{x}\:=\:{a}\:+\:\sqrt{\frac{\mathrm{4}{b}^{\mathrm{3}} \:−\:{a}^{\mathrm{3}} }{\mathrm{3}{a}}}\:\mathrm{and} \\ $$$$\mathrm{2}{y}\:=\:{a}\:−\:\sqrt{\frac{\mathrm{4}{b}^{\mathrm{3}} \:−\:{a}^{\mathrm{3}} }{\mathrm{3}{a}}}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$${x}^{\mathrm{3}} \:+\:{y}^{\mathrm{3}} \:=\:{b}^{\mathrm{3}} \:. \\ $$
Answered by esmaeil last updated on 25/Dec/23
x+y=a( i) xy=(a^2 /4)+((a^3 −4b^3 )/(12a))=((a^3 −b^3 )/(3a)) (ii) i ,ii→x^2 +y^2 =a^2 +((2b^3 −2a^3 )/(3a))=((a^3 +2b^3 )/(3a)) x^3 +y^3 = (x+y)(x^2 +y^2 −xy)= a(((a^3 +2b^3 )/(3a))+((b^3 −a^3 )/(3a)))=b^3
$${x}+{y}={a}\left(\:{i}\right) \\ $$$${xy}=\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+\frac{{a}^{\mathrm{3}} −\mathrm{4}{b}^{\mathrm{3}} }{\mathrm{12}{a}}=\frac{{a}^{\mathrm{3}} −{b}^{\mathrm{3}} }{\mathrm{3}{a}}\:\:\left({ii}\right) \\ $$$${i}\:\:,{ii}\rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} +\frac{\mathrm{2}{b}^{\mathrm{3}} −\mathrm{2}{a}^{\mathrm{3}} }{\mathrm{3}{a}}=\frac{{a}^{\mathrm{3}} +\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{3}{a}} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} = \\ $$$$\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{xy}\right)= \\ $$$${a}\left(\frac{{a}^{\mathrm{3}} +\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{3}{a}}+\frac{{b}^{\mathrm{3}} −{a}^{\mathrm{3}} }{\mathrm{3}{a}}\right)={b}^{\mathrm{3}} \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 25/Dec/23
x+y=a & x−y=(√((4b^3 − a^3 )/(3a))) 4xy=(x+y)^2 −(x−y)^2 =a^2 −((4b^3 − a^3 )/(3a))=((3a^3 +a^3 −4b^3 )/(3a)) xy=((a^3 −b^3 )/(3a)) • x^3 +y^3 =(x+y)( (x+y)^2 −3xy ) =(a)( a^2 −3(((a^3 −b^3 )/(3a))) ) =a^3 −(a^3 −b^3 )=b^3 ✓
$${x}+{y}={a}\:\&\:{x}−{y}=\sqrt{\frac{\mathrm{4}{b}^{\mathrm{3}} \:−\:{a}^{\mathrm{3}} }{\mathrm{3}{a}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}{xy}=\left({x}+{y}\right)^{\mathrm{2}} −\left({x}−{y}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:={a}^{\mathrm{2}} −\frac{\mathrm{4}{b}^{\mathrm{3}} \:−\:{a}^{\mathrm{3}} }{\mathrm{3}{a}}=\frac{\mathrm{3}{a}^{\mathrm{3}} +{a}^{\mathrm{3}} −\mathrm{4}{b}^{\mathrm{3}} }{\mathrm{3}{a}} \\ $$$${xy}=\frac{{a}^{\mathrm{3}} −{b}^{\mathrm{3}} }{\mathrm{3}{a}} \\ $$$$\bullet\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\left({x}+{y}\right)\left(\:\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{3}{xy}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\left({a}\right)\left(\:{a}^{\mathrm{2}} −\mathrm{3}\left(\frac{{a}^{\mathrm{3}} −{b}^{\mathrm{3}} }{\mathrm{3}{a}}\right)\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:={a}^{\mathrm{3}} −\left({a}^{\mathrm{3}} −{b}^{\mathrm{3}} \right)={b}^{\mathrm{3}} \:\checkmark \\ $$

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