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
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21
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.
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.
Answered by
32
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.
:)
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.
:)
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