Question

Factorize each of the following expressions: (x/2 + y + z/3)³ + (x/3 - 2y/3 + y)³ + (- 5x/6 - y/3 - 4z/3)³

Answer 1

The factorisation of $(\dfrac{x}{2}+y+\dfrac{z}{3} )^3 + (\dfrac{x}{3} -\dfrac{2y}{3}+y)^3 + (- \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3} )^3$

=$(\dfrac{x}{2}+y+\dfrac{z}{3} )(\dfrac{x}{3} -\dfrac{2y}{3}+y)(- \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3} )$

Step-by-step explanation:

The given expression:

$(\dfrac{x}{2}+y+\dfrac{z}{3} )^3 + (\dfrac{x}{3} -\dfrac{2y}{3}+y)^3 + (- \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3} )^3$

To find, the factorisation of $(\dfrac{x}{2}+y+\dfrac{z}{3} )^3 + (\dfrac{x}{3} -\dfrac{2y}{3}+y)^3 + (- \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3} )^3$ = ?

Let a = $\dfrac{x}{2}+y+\dfrac{z}{3}$, b =  $\dfrac{x}{3} -\dfrac{2y}{3}+y$ and c = $- \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3}$

∴ a + b + c

= $\dfrac{x}{2}+y+\dfrac{z}{3} + \dfrac{x}{3} -\dfrac{2y}{3}+y + - \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3}$

= $(\dfrac{x}{2}+ \dfrac{x}{3}- \dfrac{5x}{6}) +(y-\dfrac{2y}{3}+y- \dfrac{y}{3}) + (-\dfrac{4z}{3}+\dfrac{z}{3} )$

= 0

Using the algebraic identity,

If a + b + c = 0, then

$a^{3} +b^{3} +c^{3} =3abc$

∴ $(\dfrac{x}{2}+y+\dfrac{z}{3} )^3 + (\dfrac{x}{3} -\dfrac{2y}{3}+y)^3 + (- \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3} )^3$

= $(\dfrac{x}{2}+y+\dfrac{z}{3} )(\dfrac{x}{3} -\dfrac{2y}{3}+y)(- \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3} )$

Thus, the factorisation of $(\dfrac{x}{2}+y+\dfrac{z}{3} )^3 + (\dfrac{x}{3} -\dfrac{2y}{3}+y)^3 + (- \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3} )^3$

=$(\dfrac{x}{2}+y+\dfrac{z}{3} )(\dfrac{x}{3} -\dfrac{2y}{3}+y)(- \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3} )$

Answer 2

The factorisation of $(\dfrac{x}{2}+y+\dfrac{z}{3} )^3 + (\dfrac{x}{3} -\dfrac{2y}{3}+y)^3 + (- \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3} )^3$

=$(\dfrac{x}{2}+y+\dfrac{z}{3} )(\dfrac{x}{3} -\dfrac{2y}{3}+y)(- \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3} )$

Step-by-step explanation:

The given expression:

$(\dfrac{x}{2}+y+\dfrac{z}{3} )^3 + (\dfrac{x}{3} -\dfrac{2y}{3}+y)^3 + (- \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3} )^3$

To find, the factorisation of $(\dfrac{x}{2}+y+\dfrac{z}{3} )^3 + (\dfrac{x}{3} -\dfrac{2y}{3}+y)^3 + (- \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3} )^3$ = ?

Let a = $\dfrac{x}{2}+y+\dfrac{z}{3}$, b =  $\dfrac{x}{3} -\dfrac{2y}{3}+y$ and c = $- \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3}$

∴ a + b + c

= $\dfrac{x}{2}+y+\dfrac{z}{3} + \dfrac{x}{3} -\dfrac{2y}{3}+y + - \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3}$

= $(\dfrac{x}{2}+ \dfrac{x}{3}- \dfrac{5x}{6}) +(y-\dfrac{2y}{3}+y- \dfrac{y}{3}) + (-\dfrac{4z}{3}+\dfrac{z}{3} )$

= 0

Using the algebraic identity,

If a + b + c = 0, then

$a^{3} +b^{3} +c^{3} =3abc$

∴ $(\dfrac{x}{2}+y+\dfrac{z}{3} )^3 + (\dfrac{x}{3} -\dfrac{2y}{3}+y)^3 + (- \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3} )^3$

= $(\dfrac{x}{2}+y+\dfrac{z}{3} )(\dfrac{x}{3} -\dfrac{2y}{3}+y)(- \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3} )$

Thus, the factorisation of $(\dfrac{x}{2}+y+\dfrac{z}{3} )^3 + (\dfrac{x}{3} -\dfrac{2y}{3}+y)^3 + (- \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3} )^3$

=$(\dfrac{x}{2}+y+\dfrac{z}{3} )(\dfrac{x}{3} -\dfrac{2y}{3}+y)(- \dfrac{5x}{6} - \dfrac{y}{3} - \dfrac{4z}{3} )$