if a+b+c =6and ab+bc+ca=11find the value of a^3+b^3+c^3-
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Answered by
1
Answer:
we know that:
x3+y3+z3-3xyz=(x+y+z) (x2+y2+z2-xy-yz-zx)
let;
x=a, y=b, z=c
a3+b3+c3-3abc=(a+b+c)(a2+b2+c2-ab-bc-ca)
a3+b3+c3-3abc=(6)(a2+b2+c2)-(ab+bc+ca)
a3+b3+c3-3abc=6(a2+b2+c2)-(11) (OR) (OR)
a3+b3+c3-3abc=6(a+b+c)(a+b+c)-11 =6(12)-11 =6+12-11
a3+b3+c3-3abc=6x6x6-11 =72-11 =19-11
a3+b3+c3-3abc=216-11 =61 =08
a3+b3+c3-3abc=-205
Answered by
1
Answer:
Step-by-step explanation:
a³+b³+c³=(a+b+c)(a²+b²+c²-(ab+bc+ac))+3abc
a³+b³+c³-3abc=6((6)(6)-11)
a³+b³+c³-3abc=6(36-11)
=6(15)
=90
a³+b³+c³-3abc=90
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