Simplifying Algebra?

Question

1. If a+b+c=0, then a(1/b+1/c)+b(1/c+1/a)
+c(1/a+1/b)=?

2. If x=2^(1/3)-2^(-1/3), what is 2x^3+6x-3?

Answer 1

The answer to the first one is as follows:
1) multiply through & expand:a/b + a/c + b/c + b/a + c/a + c/b
2) rearrange: (a+c)/b + (a+b)/c + (b+c)/a
3) a+b+c = 0 means that a+c = -b, a+b = -c, b+c = -a
4) putting 2) and 3) together: -b/b + -c/c + -a/a
5) simplifying it: -1 + -1 + -1 = -3

Answer 2

1. If a + b + c = 0, then a(1/b+1/c) + b(1/c+1/a) + c(1/a+1/b)=?

a + b + c = 0
-a = b + c
-b = a + c
-c = a + b

a(1/b+1/c) + b(1/c+1/a) + c(1/a+1/b)
= a/b + a/c + b/c + b/a + c/a + c/b
= (b/a + c/a) + (a/b + c/b) + (a/c + b/c)
= (b + c)/a + (a + c)/b + (a + b)/c
= -a/a - b/b - c/c = -1 - 1 - 1 = -3

2. If x = 2^(1/3) - 2^(-1/3), what is 2x^3 + 6x - 3?

x³ = 2 - 3*2^(1/3) + 3*2^(-1/3) - 1/2
x³ = 3/2 - 3*2^(1/3) + 3*2^(-1/3)
2x³ = 3 - 6*2^(1/3) + 6*2^(-1/3)

6x = 6*2^(1/3) - 6*2^(-1/3)

2x³ + 6x - 3
= 3 - 6*2^(1/3) + 6*2^(-1/3) + 6*2^(1/3) - 6*2^(-1/3) - 3

2x³ + 6x - 3 = 0

Answer 3

1)
The given condition is a+b+c = 0......... (i)

a/b + a/c + b/c + b/a + c/a + c/b
=(b+c)/a + (a+c)/b + (a+b)/c
=-a/a - b/b - c/c [by (i)]
=-1 -1 -1
=-3
2)
x=2^(1/3) - 2^(-1/3)
Cubing both sides we get,
x^3 = 2 - 1/2 - 3*2^(1/3)*2^(-1/3)*{2^(1/3) - 2^(-1/3)}
or, x^3 = 3/2 - 3*1*x

[since, x=2^(1/3) - 2^(-1/3) & 2^(1/3)*2^(-1/3) = 1]

or, 2x^3 + 6x = 3
or, 2x^3 + 6x - 3 = 0

Answer 4

a(1/b+1/c) = a(1/a+1/b+1/c) - 1
b(1/c+1/a) = b(1/a+1/b+1/c) - 1
c(1/a+1/b) = c(1/a+1/b+1/c) -1

add to get LHS = (a+b+c)(1/a+1/b+1/c) -3 = 0(1/a+1/b+1/c) - 3 = - 3

2) x = 2^(1/3) - 1/2(^1/3) = a - 1/a where a= 2^(1/3)

x^3 = a^3-(1/a)^3 - 3(a-1/a)
a^3-(1/a^3) - 3x
= 2 - 1/2 - 3x
or x^3 = 3/2 - 3x
or x^3+3x - 3/2 = 0

or 2x^3+6x - 3 = 0

Answer 5

59.4