SOLUTION: Given that: a+b(1+x)^3 + c(1+2x)^3 + d(1+3x)^3 = x^3 for all values of x, find the values of the constants a,b,c and d. This is from a section of a textbook on binomial expan

Algebra ->  Permutations -> SOLUTION: Given that: a+b(1+x)^3 + c(1+2x)^3 + d(1+3x)^3 = x^3 for all values of x, find the values of the constants a,b,c and d. This is from a section of a textbook on binomial expan      Log On


   

Question 939469: Given that:
a+b(1+x)^3 + c(1+2x)^3 + d(1+3x)^3 = x^3 for all values of x, find the values of the constants a,b,c and d.
This is from a section of a textbook on binomial expansion.
The answer given is a=-1/6, b=1/2, c=-1/2 and d=1/6
I think to solve this question I have to try something like:
1^(3) + 3*1^(4) *(x) + ...but I can't see how to get the constants from this.
Any help given would be greatly appreciated.
Thanks
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
a + b(x^3 + 3x^2 + 3x+1) + c(8x^3 + 12x^2 + 6x + 1) + d(27x^3 + 27x^2 + 9x + 1) = x^3
...
3b + 12c+ + 27d = 0 ( x^2 terms)
3b + 6c + 9d = 0 (x terms)
b + 8c + 27d = 1 (x^3 terms)
b = 1/2, c = -1/2, d = 1/6
....
a + b + c + d = 0 (constant terms) a = -1/6
Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables



system%283%2Ax%2B12%2Ay%2B27%2Az=0%2C3%2Ax%2B6%2Ay%2B9%2Az=0%2C1%2Ax%2B8%2Ay%2B27%2Az=1%29



First let A=%28matrix%283%2C3%2C3%2C12%2C27%2C3%2C6%2C9%2C1%2C8%2C27%29%29. This is the matrix formed by the coefficients of the given system of equations.


Take note that the right hand values of the system are 0, 0, and 1 and they are highlighted here:




These values are important as they will be used to replace the columns of the matrix A.




Now let's calculate the the determinant of the matrix A to get abs%28A%29=-108. To save space, I'm not showing the calculations for the determinant. However, if you need help with calculating the determinant of the matrix A, check out this solver.



Notation note: abs%28A%29 denotes the determinant of the matrix A.



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Now replace the first column of A (that corresponds to the variable 'x') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5Bx%5D (since we're replacing the 'x' column so to speak).






Now compute the determinant of A%5Bx%5D to get abs%28A%5Bx%5D%29=-54. Again, as a space saver, I didn't include the calculations of the determinant. Check out this solver to see how to find this determinant.



To find the first solution, simply divide the determinant of A%5Bx%5D by the determinant of A to get: x=%28abs%28A%5Bx%5D%29%29%2F%28abs%28A%29%29=%28-54%29%2F%28-108%29=1%2F2



So the first solution is x=1%2F2




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We'll follow the same basic idea to find the other two solutions. Let's reset by letting A=%28matrix%283%2C3%2C3%2C12%2C27%2C3%2C6%2C9%2C1%2C8%2C27%29%29 again (this is the coefficient matrix).




Now replace the second column of A (that corresponds to the variable 'y') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5By%5D (since we're replacing the 'y' column in a way).






Now compute the determinant of A%5By%5D to get abs%28A%5By%5D%29=54.



To find the second solution, divide the determinant of A%5By%5D by the determinant of A to get: y=%28abs%28A%5By%5D%29%29%2F%28abs%28A%29%29=%2854%29%2F%28-108%29=-1%2F2



So the second solution is y=-1%2F2




---------------------------------------------------------





Let's reset again by letting A=%28matrix%283%2C3%2C3%2C12%2C27%2C3%2C6%2C9%2C1%2C8%2C27%29%29 which is the coefficient matrix.



Replace the third column of A (that corresponds to the variable 'z') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5Bz%5D






Now compute the determinant of A%5Bz%5D to get abs%28A%5Bz%5D%29=-18.



To find the third solution, divide the determinant of A%5Bz%5D by the determinant of A to get: z=%28abs%28A%5Bz%5D%29%29%2F%28abs%28A%29%29=%28-18%29%2F%28-108%29=1%2F6



So the third solution is z=1%2F6




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Final Answer:




So the three solutions are x=1%2F2, y=-1%2F2, and z=1%2F6 giving the ordered triple (1/2, -1/2, 1/6)




Note: there is a lot of work that is hidden in finding the determinants. Take a look at this 3x3 Determinant Solver to see how to get each determinant.