please give me the solution for this
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Answer:
8
Step-by-step explanation:
given
a^3+b^3+c^3=28
ac= 4/b
abc=4
(1/a+1/b+1/c)
by taking lcm
=(ab+bc+ca)/abc= 0(abc=4)
(so abc#0)
=> ab+bc+ca=0
a+b+c=2
from the below identity
a^3+b^3+c^3 -3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ca)
28-3*4= 2*(a^2+b^2+c^2-(ab+bc+ca))
28-12=2*(a^2+b^2+c^2-0)
16/2= a^2+b^2+c^2
a^2+b^2+c^2=8
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