Question

if a+b+c=11 and ab+ bc+.ca=2 find a^3+ b^3+c^3-3abc=?​

Answer 1

Answer:

a3+b3+c3-3abc=-205

Step-by-step explanation:

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

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Answer 2

Given, a + b + c = 6 and ab + bc + ca = 11

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 + 2ab + 2bc + 2ca

⇒ a^2 + b^2 + c^2 = (a + b + c)^2 - 2 (ab + bc + ca)

⇒ a^2 + b^2 + c^2 = (6)2 - 2 (11) = 36 - 22 = 14

⇒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 = (6) [14 - (11)]

⇒a^3 + b^3 + c^3 - 3abc = (6) [14 - (11)] = 6(3) = 18

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