find a^3+b^3+c^3-3abc if a+b+c=12 and ab+bc+ca=47
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Given a+b+c=12 and ab+bc+ca=47
Now, a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
Again, (a+b+c)^2=a^2+b^2+c^2+ab+bc+ca
⇒a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)
⇒a^2+b^2+c^2=(12)^2-2(47)=144-94=50
⇒a^3+b^3+c^3-3abc=(a+b+c)[a^2+b^2+c^2-(ab+bc+ca)]
⇒a^3+b^3+c^3-3abc=(12)[50-47]
⇒a^3+b^3+c^3-3abc=(12)(3)=36
Now, a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
Again, (a+b+c)^2=a^2+b^2+c^2+ab+bc+ca
⇒a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)
⇒a^2+b^2+c^2=(12)^2-2(47)=144-94=50
⇒a^3+b^3+c^3-3abc=(a+b+c)[a^2+b^2+c^2-(ab+bc+ca)]
⇒a^3+b^3+c^3-3abc=(12)[50-47]
⇒a^3+b^3+c^3-3abc=(12)(3)=36
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