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