How to Solve This Extreme Algebra?
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Here we’re gonna look at how to solve the following algebra problem.
Problem
From the following system of equation, find the value of a + b + c + d.
a - b + c - d = 2 ( * )
a^2 - b^2 + c^2 - d^2 = 6 ( ** )
a^3 - b^3 + c^3 - d^3 = 20 ( *** )
a^4 - b^4 + c^4 - d^4 = 66 ( **** )
Solution
The demonstrated solution might be one of various solutions out there.
First, let’s define that a + b + c + d = 2m. We’re going to apply this definition on ( * ). We use 2m simply just for mathematical convenience. Let’s take a look for the sake of clarity.
Here’s what we’ll get after subtracting the following equations.
a + b + c + d = 2m
a - b + c - d = 2
------------------ (-)
2b + 2d = 2m - 2
OR
b + d = m - 1 ( A )
On the other hand, let’s see what we’ll get after adding both equations.
a + b + c + d = 2m
a - b + c - d = 2
------------------ (+)
2a + 2c = 2m + 2
OR
a + c = m + 1 ( B )
We’re done with ( * ).
Next, we’ll do some manipulations with ( ** ).
Know that we may rewrite ( ** ) to the following form.
6 = a^2 - b^2 + c^2 - d^2
6 = a^2 + c^2 - (b^2 + d^2)
6 = (a + c)^2 - 2ac - ((b + d)^2 - 2bd)
We chose to group a with c and group b with d just because we already get the relation between a and c (a + c = m + 1), as well as b and d (b + d = m - 1).
Since we know the value of a + c and b + d in terms of m, let’s plug them in.
6 = (a + c)^2 - 2ac - ((b + d)^2 - 2bd)
6 = (m + 1)^2 - 2ac - ((m - 1)^2 - 2bd)
6 = m^2 + 2m + 1 - 2ac - (m^2 + 1 - 2m - 2bd)
6 = 4m - 2ac + 2bd
3 = 2m - ac + bd
3 - 2m = bd - ac ( C )
Alright, we finally got the relation between bd and ac from ( ** ).
Let’s proceed by working on ( *** ).
20 = a^3 - b^3 + c^3 - d^3
20 = a^3 + c^3 - (b^3 + d^3)
20 = (a + c)^3 - 3a^2.c - 3a.c^2 - ((b + d)^3 - 3b^2.d - 3b.d^2)
20 = (a + c)^3 - 3ac(a + c) - ((b + d)^3 - 3bd(b + d))
We know the value for (a + c) and (b + d) in terms of m. Let’s plug them in.
20 = (a + c)^3 - 3ac(a + c) - ((b + d)^3 - 3bd(b + d))
20 = (m + 1)^3 - 3ac(m + 1) - ((m - 1)^3 - 3bd(m - 1))
20 = m^3 + 3m^2 + 3m + 1 - 3ac(m + 1) - (m^3 - 3m^2 + 3m - 1 - 3bd(m - 1))
20 = 6m^2 + 2 - 3ac(m + 1) + 3bd(m - 1)
20 = 6m^2 + 2 - 3ac.m - 3ac + 3bd.m - 3bd
20 = 6m^2 + 2 + 3m(bd - ac) - 3(ac + bd)
Plug in the value of (bd - ac) to the above equation.
20 = 6m^2 + 2 + 3m(bd - ac) - 3(ac + bd)
20 = 6m^2 + 2 + 3m(3 - 2m) - 3(ac + bd)
20 = 6m^2 + 2 + 9m - 6m^2 - 3(ac + bd)
20 = 2 + 9m - 3(ac + bd)
18 = 9m - 3(ac + bd)
6 = 3m - (ac + bd)
3m - 6 = ac + bd ( D )
We’re done with ( *** ). Let’s proceed with ( ** ).
66 = a^4 - b^4 + c^4 - d^4
66 = a^4 + c^4 - (b^4 + d^4)
66 = (a + c)^4 - 4a^3.c - 6a^2.c^2 - 4a.c^3 - ((b + d)^4 - 4b^3.d - 6b^2.d^2 - 4b.d^3)
66 = (a + c)^4 - 2ac(2a^2 + 3ac + 2c^2) - ((b + d)^4 - 2bd(2b^2 + 3bd + 2d^2))
66 = (a + c)^4 - 2ac(2(a^2 + c^2) + 3ac) - ((b + d)^4 - 2bd(2(b^2 + d^2) + 3bd))
66 = (a + c)^4 - 2ac(2((a + c)^2 - 2ac) + 3ac) - ((b + d)^4 - 2bd(2((b + d)^2 - 2bd) + 3bd))
66 = (a + c)^4 - 2ac(2(a + c)^2 - ac) - ((b + d)^4 - 2bd(2(b + d)^2 - bd))
66 = (a + c)^4 - 4ac(a + c)^2 + 2a^2.c^2 - ((b + d)^4 - 4bd(b + d)^2 + 2b^2.d^2)
66 = (a + c)^4 - 4ac(a + c)^2 + 2a^2.c^2 - (b + d)^4 + 4bd(b + d)^2 - 2b^2.d^2
66 = (a + c)^4 - 4ac(a + c)^2 + 2(a^2.c^2 - b^2.d^2) - (b + d)^4 + 4bd(b + d)^2
66 = (a + c)^4 - 4ac(a + c)^2 + 2((ac + bd)(ac - bd)) - (b + d)^4 + 4bd(b + d)^2
Let’s plug in the value for each expression in terms of m.
66 = (m + 1)^4 - 4ac(m + 1)^2 + 2((3m - 6)(2m - 3)) - (m - 1)^4 + 4bd(m - 1)^2
66 = m^4 + 4m^3 + 6m^2 + 4m + 1 - 4ac(m^2 + 2m + 1) + 2(6m^2 - 21m + 18) - (m^4 - 4m^3 + 6m^2 - 4m + 1) + 4bd(m^2 - 2m + 1)
66 = 8m^3 + 8m - 4ac(m^2) - 8ac(m) - 4ac + 12m^2 - 42m + 36 + 4bd(m^2) - 8bd(m) + 4bd
66 = 8m^3 - 34m - 4m^2(ac - bd - 3) - 8m(ac + bd) - 4(ac - bd) + 36
66 = 8m^3 - 34m - 4m^2(2m - 6) - 8m(3m - 6) - 4(2m - 3) + 36
66 = 8m^3 - 34m - 8m^3 + 24m^2 - 24m^2 + 48m - 8m + 12 + 36
66 = 6m + 48
18 = 6m
m = 3
Since we already know that m = 3, then a + b + c + d = 2m = 2(3) = 6.