Question

PLS SEND THE 14TH QUESTION
[URGENT]

Answer 1

1.We know there's an identity

${a+b+c=0 ,then, a^3+b^3+c^3= 3abc}$

here -12+5+7=0

So

Answer will be

3×-12×5×7

= $<b>{1260}$

2. It reminds us of identity

(a+b+c)^2= a^2+b^2+c^2+2ab+2bc+2ca

here a = 2x b=3y c= -4z

so answer will be

(2x+3y-4z)^2

so factors will be

(2x+3y-4z) (2x+3y-4z)

Answer 2

14.

(i) Given (-12)^3 + (7)^3 + (5)^3.

Let x = -12, y = 7, z = 5.

x + y + z = > -12 + 7 + 5 = 0.

We know that when x + y  +z = 0, then x^3 + y^3 + z^3 = 3xyz

So, (-12)^3 + (7) + (5)

= > 3(-12)(7)(5)

$= > \boxed{1260}$

-------------------------------------------------------------------------------------------------------

(ii)

Given Equation is 4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz

= > (2x)^2 + (3y)^2 + (4z)^2 + 2(2x)(3y) - 2(3y)(4z) - 2(2x)(4z)

It is in the form of a^2 + b^2 + c^2 + 2ab - 2bc - 2ca = (a + b - c)^2

$= > \boxed{(2x + 3y - 4z)^2}$

Hope this helps!