Question

1. If x+y+z=0, then show that x^2+y^2+z^2=-2(xy+yz+zx).
7

2. If a=3+b, prove that a^3-b^3-9ab-27.

3. Factorise 9 x^2-12x+4.

4. Factorise 343+27t^3

Please...
I will give you 50 points
Don't cheat​

Answer 1

1. x + y + z = 0

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

or (0)² = x² + y² + z² + 2xy + 2yz + 2zx

or 0 = x² + y² + z² + 2xy + 2yz + 2zx

or 0 - 2xy - 2yz - 2zx = x² + y² + z²

or x² + y² + z² = -2xy - 2yz - 2zx

or x² + y² + z² = -2(xy + yz + zx)

2. Question is incomplete!!!

3. 9x² - 12x + 4

= (3x)² - 2(3x)(2) + (2)²

= (3x - 2)²

= (3x -2)(3x - 2).

4. 343 + 27t³

= (7)³ + (3t)³

= (7 + 3t){(7)² - (7)(3t) + (3t)²}

= (7 + 3t)(49 - 21t + 9t²).

Identity used above....

a³ + b³ = (a+b)(a²-ab+b²).

Answer 2

Answer:-

1] x + y + z = 0

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

or (0)² = x² + y² + z² + 2xy + 2yz + 2zx

or0 = x² + y² + z² + 2xy + 2yz + 2zx

or0 - 2xy - 2yz - 2zx = x² + y² + z²

orx² + y² + z² = -2xy - 2yz - 2zx

or x² + y² + z² = -2(xy + yz + zx)

2] a= 3+b ( given).

a-b=3.

take cube both sides.

(a-b)³ = (3)³

a³ - b³ -3ab * 3 = 27.

a³ - b³ - 9 ab = 27.

3] 9x² - 12x + 4

= (3x)² - 2(3x)(2) + (2)²

= (3x - 2)²

= (3x -2)(3x - 2).

4] 343 + 27t³

= (7)³ + (3t)³

= (7 + 3t){(7)² - (7)(3t) + (3t)²}

= (7 + 3t)(49 - 21t + 9t²).

identity used

a³ + b³ = (a+b)(a²-ab+b²).

i hope it helps you.