Question

Factorise: (3x - 1/2 y + 1/3 z)^2​

Answer 1

Given:

$(3x - \frac{1}{2} y + \frac{1}{3} z)^{2}$

Taking the LCM as 6,

$( \frac{1}{6} (18x−3y+2z))^{2}$

Applying the product rule to $\frac{1}{6} (18x−3y+2z)^{2}$:-

$= > ( \frac{1}{6})^{2} (18x−3y+2z)^{2}$

Now applying the product rule to 1/6:-

$\frac{1^{2} }{ {6}^{2} }(18x - 3y + 2z)^{2}$

Hence, the answer.

Answer 2

Step-by-step explanation:

Given expression is {3x - (1/2)y + (1/3)z}²

⇛{3x + (-1/2)y + (1/3)z}²

Now,

This is in the form of (a+b+c)²

Here,

a = 3x, b = (-1/2)y and c = (1/3)z

So, using identity (a+b+c)² = a²+b²+c²+2ab+2bc+2ca, we get

{3x + (-1/2)y + (1/3)z}²

= (3x)²+{-(1/2)y}² + {(1/3)z}² + 2(3x){-(1/2)y} + 2{-(1/2)y}{(1/3)z} + 2{(1/3)z}(3x)

= 3*3*x*x+ {(-1*-1/2*2)y*y} + {(1*1/3*3)z*z} + 2(3x){-(1/2)y} + 2{-(1/2)y}{(1/3)z} + 2{(1/3)z}(3x - (yz/3) + 2zx

= 9x² + (y²/4) + (z²/9) - 3xy - (yz/3) + 2zx

Take the LCM of the denominator 4, 3 and 9 is 36.

= {(32x² + 9y² + 4z² - 108xy - 12yz + 72zx)/36}

= (1/35)(32x² + 9y² + 4z² - 108xy - 12yz + 72zx)

Answer: Therefore, {3x - (1/2)y + (1/3)z}² = (1/35)(32x² + 9y² + 4z² - 108xy - 12yz + 72zx)

Please let me know if you have any other.