Using given determinant eqn, prove abc(ab+bc+ca) = a+b+c

[emma246000]

Using the determinant given below: (which is equal to zero)

| a a^3 a^4-1 |
| b b^3 b^4-1 | = 0
| c c^3 c^4-1 |

...prove that abc (ab+bc+ca) = a+b+c

Here is what I know: We can split this determinant up as follows:

| a a^3 a^4 |   | a a^3 1 |
| b b^3 b^4 | - | b b^3 1 |
| c c^3 c^4 |   | c c^3 1 |

After this step I am totally clueless and really confused!

 

[stapel]

Since:

. . . . .\(\displaystyle \L \left| \begin{array}{ccc}a&a^3&a^4\\b&b^3&b^4\\c&c^3&c^4\end{array}\right| \,-\, \left| \begin{array}{ccc}a&a^3&1\\b&b^3&1\\c&c^3&1\end{array} \right|\)

...does not, in general, equal:

. . . . .\(\displaystyle \L \left| \begin{array}{ccc}a&a^3&a^4-1\\b&b^3&b^4-1\\c&c^3&c^4-1\end{array} \right|\)

...I'm not sure on what basis you feel you can make this substitution...?

Note: In general, for the matrix sum A + B = C, det(A) + det(B) is NOT equal to det(C) = det(A + B).

Instead, why not try finding the determinant? Find the "value" by the usual method, set the resulting expression equal to zero, and see if you can rearrange the resulting equation into the desired form.

Eliz.