Simplifying Determinants to Finding Linear Factors - Homework Help

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The discussion revolves around expressing a given determinant as a product of four linear factors. Participants struggle with manipulating the determinant and finding a clear pattern for factorization. Suggestions include row reducing the matrix and expanding along the first row to simplify the expression. The identity for the difference of cubes, a^3 - b^3 = (a - b)(a^2 + ab + b^2), is highlighted as a useful tool in the process. Ultimately, this identity aids in making progress toward the solution.
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Homework Statement



Express the determinant as a product of four linear factors.

<br /> \left(<br /> \begin{array}{ccc}<br /> 1 &amp; a &amp; a^3 \\<br /> 1 &amp; b &amp; b^3 \\<br /> 1 &amp; c &amp; c^3<br /> \end{array}<br /> \right)<br />

I'm sure that the only way to do this without hurting yourself is to operate on the determinant and take factors out. It just doesn't move. I can't see the pattern, I can't see how to get factors out of it. I add rows, columns everything.
 
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Expand along the first row and then just try to factorize it.
 
rock.freak667 said:
Expand along the first row and then just try to factorize it.

Can't do it. It's hard to show my working on latex buts there's pages of attempts. x^3 I just can't do.
 
Try row reducing once, with R2-R1 and R3-R1. Then expand along row one.


then use this identity and see if it helps a3-b3=(a-b)(a2+ab+b2)
 
rock.freak667 said:
Try row reducing once, with R2-R1 and R3-R1. Then expand along row one.


then use this identity and see if it helps a3-b3=(a-b)(a2+ab+b2)


Excellent thanks, that's the identity I needed.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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