Question

If a+b+c=17 and a2 + b2 + c2,=89 ,find the value of a3 +b3 + c3 -3abc.​

Answer 1

Answer:

(a+b+c)^3=4913(a+b+c)

3

=4913

Step-by-step explanation:

Given : If a^2+b^2 + c^2= 280a

2

+b

2

+c

2

=280 and ab+bc+ca=\frac{9}{2}ab+bc+ca=

2

9

To find : The value of (a+b+c)^3(a+b+c)

3

Solution :

We have given,

a^2+b^2 + c^2= 280a

2

+b

2

+c

2

=280

Expand using identity, a^2+b^2 + c^2=(a+b+c)^2-2(ab+bc+ca)a

2

+b

2

+c

2

=(a+b+c)

2

−2(ab+bc+ca)

(a+b+c)^2-2(ab+bc+ca)= 280(a+b+c)

2

−2(ab+bc+ca)=280

(a+b+c)^2-2(\frac{9}{2})= 280(a+b+c)

2

−2(

2

9

)=280

(a+b+c)^2-9= 280(a+b+c)

2

−9=280

(a+b+c)^2= 280+9(a+b+c)

2

=280+9

(a+b+c)^2= 289(a+b+c)

2

=289

Taking square root both side,

(a+b+c)= \sqrt{289}(a+b+c)=

289

(a+b+c)=17(a+b+c)=17

Taking cube both side,

(a+b+c)^3=(17)^3(a+b+c)

3

=(17)

3

(a+b+c)^3=4913(a+b+c)

3

=4913

Therefore, (a+b+c)^3=4913(a+b+c)

3

=4913