if a+b+c=13 and a^2+b^2+c^2=83 then find the value of a^3+b^3+c^3-3abc
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Step-by-step explanation:
Given: a + b + c = 15, a² + b² + c² = 83
∴ (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
⇒ 15² = 83 + 2(ab + bc + ca)
⇒ 225 - 83 = 2(ab + bc + ca)
⇒ 142 = 2(ab + bc + ca)
⇒ ab + bc + ca = 71
Now,
a³ + b³ + c³ - 3abc:
= (a + b + c)(a² + b² + c² - ab - bc - ca)
= (a + b + c)(a² + b² + c² - (ab + bc + ca))
= (15)(83 - 71)
= 180.
Hence, the value of a³+b³+c³-3abc is 180
Hope it helps!
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