Question

if a+b+c=6 and a^2+b^2+c^2=38 then find the value of a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)+3abc​

Answer 1

Step-by-step explanation:

first find abc and substitute

Answer 2

Let us look at the answer:

The answer is -6.

Step-by-step explanation:

a+b+c = 6 ...... (i)   and a²+b²+c²= 38 .... (ii)

squaring both sides of ..(i) (a²+b²+c²+2ab+2bc+2ca) = (a+b+c)²

                                           38+2(ab+bc+ca) = 36

                                           2(ab+bc+ca) = -2  ⇒ ab+bc+ca = -1

                                       

From .. (ii) b²+c²= 38-a² , a²+b²= 38-c², b²+ c²= 38-b²

We need to find the value of a(b²+c²)+ b(c²+a²) + c( a²+ b²) +3abc

∴ a(b²+c²)+ b(c²+a²) + c( a²+ b²) +3abc = a(38-a²)+ b(38-b²) +c(38-c²) +3abc

                                                             = -(a³+b³+c³) +38(a+b+c) +3abc  

                                                        = -( a³+b³+c³-3abc) +38(a+b+c)            ....(iii)

So, we need to find the value of a³+b³+c³-3abc

Using the identity, a³+b³+c³-3abc = (a+b+c) (a²+b²+c²-ab-bc-ac)  

so, a³+b³+c³-3abc = (6) {38-(-1)} = 6× 39 = 234

substituting the values in (iii) ,

a(b²+c²)+ b(c²+a²) + c( a²+ b²) +3abc = -234 +38×6 = -234+228 = -6