Question

if ab+bc+ca=9 and a²+b²+c²=18 find the vlaue of 3(a+b+c)​

Answer 1

Answer:

18

Step-by-step explanation:

(a+b+c)^2=$a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$

putting given values we get $(a+b+c)^{2}$=36

so a+b+c=6

therefore 3(a+b+c)=18

Answer 2

Answer:

18

Step-by-step explanation:

(a+b+c)² = a²+b²+c²+2ab+2bc+2ca

(a+b+c)² = (a²+b²+c²)+2(ab+bc+ca)

(a+b+c)² = 18+(2×9)

(a+b+c)² = 18+18

(a+b+c)² = 36

(a+b+c) = √36

(a+b+c) = 6

3(a+b+c) = 3×6 = 18