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$$\begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ a^3 & b^3 & c^3 \end{vmatrix} = (a-b)(b-c)(c-a)(a+b+c)$$

we have to solve this by using the properties of determinants without actually expanding the determinant. I am Unable to think which calculation to apply so was hoping for an hint.

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  • $\begingroup$ oops, fixed typo $\endgroup$ – Raza Hassan Aug 30 '14 at 16:38
  • $\begingroup$ what properties of determinants are you allowed to use? $\endgroup$ – djechlin Aug 30 '14 at 21:48
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Using $C_2'=C_2-C_1, C_3'=C_3-C_1$

$$\begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ a^3 & b^3 & c^3 \end{vmatrix}$$

$$=\begin{vmatrix} 1 & 0 & 0\\ a & b-a & c-a\\ a^3 & b^3-a^3 & c^3-a^3 \end{vmatrix}$$

$$=-(a-b)(c-a)\begin{vmatrix} 1 & 0 & 0\\ a & 1 & 1\\ a^3 & b^2+ab+a^2 & c^2+ca+a^2 \end{vmatrix}$$

Use $C_2'=C_2-C_3$

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say, $ \begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ a^3 & b^3 & c^3 \end{vmatrix} = f(a,b,c) $

Clearly, when $a=b$, the first two columns are identical making $f = 0 $
So $(a-b)$ is a factor of $f$ by factor theorem

Similarly, $(b-c)$ and $(c-a)$ are also factors of $f$ :
$$f(a,b,c) = (a-b)(b-c)(c-a)*g(a,b,c)$$

See if you can think of a way to find out $g(a,b,c)$

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Apart from Ganeshie8's, these answers are longer than just expanding the determinant. The conceptual answer is to say the determinant is an alternating degree 4 polynomial in $a,b,c$ with all coefficients 1. So it is the Vandermonde polynomial on $a,b,c$ times the symmetric degree 1 polynomial in them with coefficients 1, which is $a+b+c$. See http://en.wikipedia.org/wiki/Vandermonde_polynomial

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HINT

Subtract 1st column from second than 2nd from 3rd

apply identity of a^3-b^3 and take a-b and b-c common from 1st two columns then subtract 1st from 2nd and then open it at last .

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