Question

if a+b+c=6 and a2+b2+c2=60,then ab+bc+ca and a3+b3+c3-3abc.​

Answer 1

explanation is in the picture.

Hope this helps you

Answer 2

The required value is -12 and 216.

Step-by-step explanation:

Since we have given that

$a+b+c=6$

$a^2+b^2+c^2=60$

Now, squaring both sides, we get that

$(a+b+c)^2=6^2\\\\a^2+b^2+c^2+2ab+2bc+2ca=36\\\\60+2(ab+bc+ca)=36\\\\2(ab+bc+ca)=36-60=-24\\\\ab+bc+ca=-12$

So,

$a^3+b^3+c^3-3abc=\dfrac{1}{2}(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\\\\=0.5(6)(60-(-12))\\\\=0.5(6)(60+12)\\\\=3\times 72\\\\=216$

Hence, the required value is -12 and 216.

# learn more:

(a3+b3 +c3-3abc) /a2+b2+c2-ab-bc-ca)

https://brainly.in/question/4956100