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Question Number 190115 by HeferH last updated on 27/Mar/23
if:  (a+b)(a+1) = b   find:  P =  (√(a^3 +b^3 −3ab))
$$\mathrm{if}:\:\:\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}+\mathrm{1}\right)\:=\:\mathrm{b} \\ $$$$\:\mathrm{find}:\:\:\mathrm{P}\:=\:\:\sqrt{\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} −\mathrm{3ab}} \\ $$
Answered by som(math1967) last updated on 27/Mar/23
(a+b)(a+1)=b  ⇒a^2 +a+ab+b=b  ⇒a(a+b+1)=0  if a=0  then P=(√b^3 )=b^(2/3)   if a+b+1=0  (a+b)=−1  ⇒(a+b)^3 =−1  a^3 +b^3 +3ab(a+b)=−1  ⇒ a^3 +b^3 −3ab=−1  ⇒(√(a^3 +b^3 −3ab))=(√(−1))  ∴P=i
$$\left({a}+{b}\right)\left({a}+\mathrm{1}\right)={b} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{a}+{ab}+{b}={b} \\ $$$$\Rightarrow{a}\left({a}+{b}+\mathrm{1}\right)=\mathrm{0} \\ $$$${if}\:{a}=\mathrm{0} \\ $$$${then}\:{P}=\sqrt{{b}^{\mathrm{3}} }={b}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$${if}\:{a}+{b}+\mathrm{1}=\mathrm{0} \\ $$$$\left({a}+{b}\right)=−\mathrm{1} \\ $$$$\Rightarrow\left({a}+{b}\right)^{\mathrm{3}} =−\mathrm{1} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +\mathrm{3}{ab}\left({a}+{b}\right)=−\mathrm{1} \\ $$$$\Rightarrow\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} −\mathrm{3}{ab}=−\mathrm{1} \\ $$$$\Rightarrow\sqrt{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} −\mathrm{3}{ab}}=\sqrt{−\mathrm{1}} \\ $$$$\therefore{P}={i} \\ $$
Commented by HeferH last updated on 27/Mar/23
Thanks!
$$\mathrm{Thanks}! \\ $$

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