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, \[ \frac{a^3+b^3+c^3}{abc} = \frac{3abc}{abc} \]

Find the Value of x^3 + y^3 – 12xy + 64 Find the value of : \( x^3 + y^3 – 12xy + 64 \) when \( x + y = -4 \) \[ x^3 + y^3 – 12xy + 64 \] \[ = x^3 + y^3 + (-4)^3 – 3xy(-4) \] \[ = x^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 (x – 2y)^3 + (2y – 3z)^3 + (3z – x)^3 Factorize the following expression : \( (x – 2y)^3 + (2y – 3z)^3 + (3z – x)^3 \) \[ (x – 2y)^3 + (2y – 3z)^3 + (3z – x)^3 \] \[ \text{Let } a = (x – 2y),\quad b = (2y –

Factorize (a – 3b)^3 + (3b – c)^3 + (c – a)^3 Factorize the following expression : \( (a – 3b)^3 + (3b – c)^3 + (c – a)^3 \) \[ (a – 3b)^3 + (3b – c)^3 + (c – a)^3 \] \[ \text{Let } x = (a – 3b),\quad y = (3b –

Factorize (2x – 3y)^3 + (4z – 2x)^3 + (3y – 4z)^3 Factorize the following expression : \( (2x – 3y)^3 + (4z – 2x)^3 + (3y – 4z)^3 \) \[ (2x – 3y)^3 + (4z – 2x)^3 + (3y – 4z)^3 \] \[ \text{Let } a = (2x – 3y),\quad b = (4z –

Factorize (3x – 2y)^3 + (2y – 4z)^3 + (4z – 3x)^3 Factorize the following expression : \( (3x – 2y)^3 + (2y – 4z)^3 + (4z – 3x)^3 \) \[ (3x – 2y)^3 + (2y – 4z)^3 + (4z – 3x)^3 \] \[ \text{Let } a = (3x – 2y),\quad b = (2y –

Find the Value of 48^3 – 30^3 – 18^3 Without actually computing the cubes, find the value of : \( 48^3 – 30^3 – 18^3 \) \[ 48^3 – 30^3 – 18^3 \] \[ = 48^3 + (-30)^3 + (-18)^3 \] \[ \text{Since } 48 + (-30) + (-18) = 0 \] \[ \therefore a^3

Factorize 2√2a^3 + 16√2b^3 + c^3 – 12abc Factorize the following expression : \( 2\sqrt{2}a^3 + 16\sqrt{2}b^3 + c^3 – 12abc \) \[ 2\sqrt{2}a^3 + 16\sqrt{2}b^3 + c^3 – 12abc \] \[ = (\sqrt{2}a)^3 + (2\sqrt{2}b)^3 + c^3 – 3(\sqrt{2}a)(2\sqrt{2}b)(c) \] \[ = (\sqrt{2}a + 2\sqrt{2}b + c) \] \[ \times \left[ 2a^2 + 8b^2

Factorize 3√3a^3 – b^3 – 5√5c^3 – 3√15abc Factorize the following expression : \( 3\sqrt{3}a^3 – b^3 – 5\sqrt{5}c^3 – 3\sqrt{15}abc \) \[ 3\sqrt{3}a^3 – b^3 – 5\sqrt{5}c^3 – 3\sqrt{15}abc \] \[ = (\sqrt{3}a)^3 + (-b)^3 + (-\sqrt{5}c)^3 – 3(\sqrt{3}a)(-b)(-\sqrt{5}c) \] \[ = (\sqrt{3}a – b – \sqrt{5}c) \] \[ \times \left[ 3a^2 + b^2