Forgot password? New user? Sign up
Existing user? Log in
{a+b+c=−abc1a+1b+1c=1\large \begin{cases} a + b + c = -abc \\ \dfrac 1a + \dfrac 1b + \dfrac 1c = 1 \end{cases} ⎩⎪⎨⎪⎧a+b+c=−abca1+b1+c1=1
Given the above, where a,b,c≠0a, b, c \ne 0a,b,c=0, find 1a2+1b2+1c2\dfrac 1{a^2} + \dfrac 1{b^2} + \dfrac 1{c^2}a21+b21+c21.
Problem Loading...
Note Loading...
Set Loading...