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Use identity,
![<br />a^3+b^3+c^3-3abc\\<br />=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\\<br />=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]\\<br />=4[4^2-3\times6]\\<br />=4\times(-2)=-8<br />](https://tex.z-dn.net/?f=%3Cbr+%2F%3Ea%5E3%2Bb%5E3%2Bc%5E3-3abc%5C%5C%3Cbr+%2F%3E%3D%28a%2Bb%2Bc%29%28a%5E2%2Bb%5E2%2Bc%5E2-ab-bc-ca%29%5C%5C%3Cbr+%2F%3E%3D%28a%2Bb%2Bc%29%5B%28a%2Bb%2Bc%29%5E2-3%28ab%2Bbc%2Bca%29%5D%5C%5C%3Cbr+%2F%3E%3D4%5B4%5E2-3%5Ctimes6%5D%5C%5C%3Cbr+%2F%3E%3D4%5Ctimes%28-2%29%3D-8%3Cbr+%2F%3E "<br />a^3+b^3+c^3-3abc\\<br />=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\\<br />=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]\\<br />=4[4^2-3\times6]\\<br />=4\times(-2)=-8<br />")
Answered by
1
by formula
a3+b3+c3-3abc= (a+b+c)(a2+b2+c2-ab-bc-ca)
u need the value of a2+b2+c2
a2+b2+c2=(a+b+c)squ-2(ab+bc+ca)
hence a2+b2+c2=4
now putting value
4(4-6)
=4×-2
=-8
hence proved
a3+b3+c3-3abc= (a+b+c)(a2+b2+c2-ab-bc-ca)
u need the value of a2+b2+c2
a2+b2+c2=(a+b+c)squ-2(ab+bc+ca)
hence a2+b2+c2=4
now putting value
4(4-6)
=4×-2
=-8
hence proved
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