Question

if a + b + c is equals to 6 and a b + BC + CA is equals to 11 find a cube plus b cube plus c cube minus 3 ABC

Answer 1

. if it helps mark as brainliest

Answer 2

Answer: The answer is 18.

Step-by-step explanation:  Given that

$a+b+c=6,\\\\ab+bc+ca=11,\\\\a^3+b^3+c^3-3abc=?$

We know the following factorisation formula

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).$

Now,

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\\\\\Rightarrow a^2+b^2+c^2=(a+b+c)^2-2ab-2bc-2ca$

So, we have

$a^3+b^3+c^3-3abc\\\\=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\\\\=(a+b+c)\{(a+b+c)^2-2ab-2bc-2ca-ab-bc-ca\}\\\\=(a+b+c)\{(a+b+c)^2-3(ab+bc+ca)\}\\\\=6\times\{6^2-3\times 11\}\\\\=6\times (36-33)\\\\=6\times 3\\\\=18.$.

Thus, the answer is 18.