Question

Question for Brainly Teachers​

Answer 1

Step-by-step explanation:

if a,b and C are all different

$\left|\begin{array}{ccc}</p><p>a&a^2&a^4-1\\</p><p>b&b^2&b^4-1\\</p><p>c&c^2&c^4-1\end{array}\right|=0$

then prove that

$ab+bc+ca=\frac{a+b+c}{abc}$

Separate the determinant

$\left|\begin{array}{ccc}</p><p>a&a^2&a^4\\</p><p>b&b^2&b^4\\</p><p>c&c^2&c^4\end{array}\right|+\left|\begin{array}{ccc}</p><p>a&a^2&-1\\</p><p>b&b^2&-1\\</p><p>c&c^2&-1\end{array}\right|=0$

Take a,b,c common from R1,R2 and R3 respectively from 1st determinant

$abc\left|\begin{array}{ccc}</p><p>1&a&a^3\\</p><p>1&b&b^3\\</p><p>1&c&c^3\end{array}\right|-\left|\begin{array}{ccc}</p><p>a&a^2&1\\</p><p>b&b^2&1\\</p><p>c&c^2&1\end{array}\right|=0$

Apply R2 -> R2-R1

and R3 -> R3-R1

in both

$abc\left|\begin{array}{ccc}</p><p>1&a&a^3\\</p><p>1-1&b-a&b^3-a^3\\</p><p>1-1&c-a&c^3-a^3\end{array}\right|-\left|\begin{array}{ccc}</p><p>a&a^2&1\\</p><p>b-a&b^2-a^2&1-1\\</p><p>c-a&c^2-a^2&1-1\end{array}\right|=0$

$abc\left|\begin{array}{ccc}</p><p>1&a&a^3\\</p><p>0&b-a&(b-a)(b^2+ab+a^2)\\</p><p>0&c-a&(c-a)(c^2+ac+a^2)\end{array}\right|-\left|\begin{array}{ccc}</p><p>a&a^2&1\\</p><p>b-a&(b-a)(b+a)&0\\</p><p>c-a&(c-a)(c+a)&0\end{array}\right|=0$

take common (b-a) and (c-a) from R2 and R3 in both Determinants

$abc(b-a)(c-a)\left|\begin{array}{ccc}</p><p>1&a&a^3\\</p><p>0&1&(b^2+ab+a^2)\\</p><p>0&1&(c^2+ac+a^2)\end{array}\right|-(b-a)(c-a)\left|\begin{array}{ccc}</p><p>a&a^2&1\\</p><p>1&(b+a)&0\\</p><p>1&(c+a)&0\end{array}\right|=0$

Expanding 1st determinant along C1 and second by C3

$abc(b-a)(c-a)\left|\begin{array}{cc}</p><p>1&(b^2+ab+a^2)\\</p><p>1&(c^2+ac+a^2)\end{array}\right|-(b-a)(c-a)\left|\begin{array}{cc}</p><p>1&(b+a)\\</p><p>1&(c+a)\end{array}\right|=0$

$abc(b-a)(c-a)[c^2+ac+a^2-b^2-ab-a^2]-(b-a)(c-a)[c+a-b-a]=0\\\\abc(b-a)(c-a)[c^2+ac-b^2-ab]-(b-a)(c-a)[c-b]=0$

$abc(b-a)(c-a)(c-b)[c+b+a]-(b-a)(c-a)[c-b]=0$$(b-a)(c-a)(c-b)[abc(a+b+c)-1]=0$

$[abc(a+b+c)-1]=0$$abc(a+b+c)=1$

Note*:From the given data the above mentioned can be proved.

If the data was

if a,b and C are all different

$\left|\begin{array}{ccc}</p><p>a&a^3&a^4-1\\</p><p>b&b^3&b^4-1\\</p><p>c&c^3&c^4-1\end{array}\right|=0$

then only we can prove that

$ab+bc+ca=\frac{a+b+c}{abc}$

For this do the same process,and you will reach the results.

Hope it helps you