Question

if a + b + c = 10 and a b + BC + AC = 31 find the value of a cube plus b cube plus c cube minus 3 ABC

Answer 1

Given that,
a+b+c=10.................................................1
and,
ab+bc+ac=31...........................................2
Squaring on both sides to eq1 we have,
(a+b+c)^2=10^2

a^2+b^2+c^2+2(ab+bc+ac)=100

a^2+b^2+c^2+2×31=100 (by eq2)

a^2+b^2+c^2=38....................................3
We know that,
a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)
By eq1 eq2 and eq3 we have,
a^3+b^3+c^3-3abc=(10)(38-31)=70

Hence a^3+b^3+c^3-3abc=70.

Answer 2

Given,
a+b+c = 10
ab + bc + ca = 31

a+b+c = 10

(a+b+c)² = 10² ( Squaring on both sides)

a²+b²+c²+2ab+2bc+2ca = 100

a²+b²+c² +2(ab+bc+ca) = 100

a²+b²+c²+2(31) = 100 ( ab+bc+ca) = 31

a²+b²+c² = 100-62

a²+b²+c² = 38.


We know that,
a³+b³+c³-3abc = (a+b+c) X (a2+b2+c2-ab-bc-ca)

= [a+b+c] [a²+b²+c² - ( ab+bc+ca )

= 10 [ 38 - ( 31 ) ]

= 10 [ 7 ]

= 70.

:)