Question

if a+b+c=15,ab+bc+ca=74 & abc=120. find a cube+b cube+c cube​

Answer 1

Answer:

a^3+b^3+c^3 = 405.

Step-by-step explanation:

given

➡a+b+c =15

➡ab+bc+ca =74

➡abc =120

find

a^3+b^3+c^3=?

solution

from the formula ..

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

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

(15)^2=a^2 + b^2 + c^2 + 2 (74)

225 = a^2 + b^2 + c^2 + 148

a^2 + b^2 + c^2 = 225 - 148

a^2 + b^2 + c^2 = 77.

from formula..

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

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

(15) (77-(74))+3(120) = a^3+b^3+c^3

a^3+b^3+c^3 = (15×3)+360 = 45+360

a^3+b^3+c^3 = 405.