If a, b, c are all non-zero and a + b + c = 0, prove that a^2/bc + b^2/ca + c^2/ab = 3.

Prove that a²/bc + b²/ca + c²/ab = 3 If \(a, b, c\) are all non-zero and \(a+b+c=0\), prove that \[ \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}=3 \] \[ \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab} \] \[ = \frac{a^3+b^3+c^3}{abc} \] Since, \[ a+b+c=0 \] Using the identity \[ a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \] Therefore, \[ a^3+b^3+c^3-3abc=0 \] \[ a^3+b^3+c^3=3abc \] Hence, \[ \frac{a^3+b^3+c^3}{abc} = \frac{3abc}{abc} \]

If a, b, c are all non-zero and a + b + c = 0, prove that a^2/bc + b^2/ca + c^2/ab = 3. Read More »

Factorize the following expression : (x/2 + y + z/3)^3 + (x/3 – 2y/3 + z)^3 + (-5x/6 – y/3 – 4z/3)^3

Factorize (x/2 + y + z/3)^3 + (x/3 – 2y/3 + z)^3 + (-5x/6 – y/3 – 4z/3)^3 Factorize the following expression : \[ \left(\frac{x}{2}+y+\frac{z}{3}\right)^3 + \left(\frac{x}{3}-\frac{2y}{3}+z\right)^3 + \left(-\frac{5x}{6}-\frac{y}{3}-\frac{4z}{3}\right)^3 \] \[ \text{Let } a=\left(\frac{x}{2}+y+\frac{z}{3}\right), \] \[ b=\left(\frac{x}{3}-\frac{2y}{3}+z\right), \] \[ c=\left(-\frac{5x}{6}-\frac{y}{3}-\frac{4z}{3}\right) \] \[ a+b+c=0 \] \[ \therefore a^3+b^3+c^3=3abc \] \[ =3\left(\frac{x}{2}+y+\frac{z}{3}\right) \left(\frac{x}{3}-\frac{2y}{3}+z\right) \left(-\frac{5x}{6}-\frac{y}{3}-\frac{4z}{3}\right) \] Next Question

Factorize the following expression : (x/2 + y + z/3)^3 + (x/3 – 2y/3 + z)^3 + (-5x/6 – y/3 – 4z/3)^3 Read More »