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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))
if:(a+b)(a+1)=bfind:P=a3+b3−3ab
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
(a+b)(a+1)=b⇒a2+a+ab+b=b⇒a(a+b+1)=0ifa=0thenP=b3=b23ifa+b+1=0(a+b)=−1⇒(a+b)3=−1a3+b3+3ab(a+b)=−1⇒a3+b3−3ab=−1⇒a3+b3−3ab=−1∴P=i
Commented by HeferH last updated on 27/Mar/23
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