Simultaneous Equations
Given the following simultaneous equations, determine the product a b c .
a
3
+
1
1
(
b
2
+
c
2
)
+
3
6
a
=
5
7
5
,
b
3
+
1
1
(
c
2
+
a
2
)
+
3
6
b
=
5
7
5
,
c
3
+
1
1
(
a
2
+
b
2
)
+
3
6
c
=
5
7
5
.
The answer is 36.
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2 solutions
Please use LaTeX
?? can you explain more clearly?? Thanx
In LaTeX form (with few minor edits):
Let P 2 = a 2 + b 2 + c 2 , then each of a , b , c are the roots to x 3 − 1 1 x 2 + 3 6 x = 5 7 5 − 1 1 P 2 = 5 7 5 − 1 1 ( s 1 2 − 2 s 2 ) where s 1 = a + b + c and s 2 = a b + b c + c a . Then by Vieta's we have s 1 = 1 1 and s 2 = 3 6 and consequently 5 7 5 − 1 1 ( s 1 2 − 2 s 2 ) = 5 7 5 − 1 1 ( 1 1 2 − 2 ∗ 3 6 ) = 5 7 5 − 1 1 ( 4 9 ) = 3 6 . Our given equation is now reduced to x 3 − 1 1 x 2 + 3 6 x − 3 6 = 0 . The product of the roots, which is also a b c , is thus equal to 3 6
By adding the equations be get:- a 3 + b 3 + c 3 + 2 2 ( a 2 + b 2 + c 2 ) + 3 6 ( a + b + c ) = 3 × 5 7 5
Comparing with newton's identity:- a 4 + b 4 + c 4 = ( a + b + c ) ( a 3 + b 3 + c 3 ) − ( a b + b c + a c ) ( a 2 + b 2 + c 2 ) + a b c ( a + b + c ) a b c = 3 6
Let P2=a^2+b^2+c^2, then each of a, b, c are the roots to X^3+11(P2-x^2)+36x=575 P2=(s1^2-2s2) where s1=a+b+c and s2=ab+bc+ca x^3-11x^2+36x-575+11(s1^2-2s2)=0 Then by vieta s1=11 and s2=36. We are looking for abc which is s3 which is -575+11(s1^2-2s2) which works out to -36 so answer is 36.