Math, asked by kb80101nb, 8 months ago

if a+b+c =6and ab+bc+ca=11find the value of a^3+b^3+c^3-​

Answers

Answered by Raghavrocks17
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 maddulamounika111
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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