x³+(2y)³+(-1)³-3(x)(2y)(-1)
= (x+2y-1)(x²+4y²+1-2xy+2y+x)

m³+512 = m³+8³
a³+b³ = (a+b)(a²-ab+b²)
(m+8) (m²-8m+64)

Given, x-y = 4, xy = 5
x³-y³= (x-y)³+3xy(x-y)
= 4³+3(5)(4)
=124

36m²-49n² = (6m)²-(7n)²
= (6m+7n)(6m-7n)  [∵ a²-b²=(a+b)(a-b)]

(2a-3b+4c)²= (2a+(-3b)+4c)² = (2a)²+(-3b)²+(4c)²+2(2a)(-3b)+2(-3b)(4c)+2(4c)(2a)
= 4a²+9b²+16c²+(-12ab)+(-24bc)+16ca
= 4a²+9b²+16c²-12ab-24bc+16ca

We know that, (x+y)³ = x³+3x²y+3xy²+y³
(2x+3y)³ = (2x)³+3(2x)²(3y)+3(2x)(3y)²+(3y)³
= 8x³+3(4x²)(3y)+3(2x)(9y²)+27y³
= 8x³+36x²y+54xy²+27y³

Let P(x) = kx³-2x²+25x-26
By factor theorem, (x - 1) divides P(x) without remainder, P (1) = 0
P(1) k(1)³ -2 (1)² + 25 (1) - 26 = 0
k-2+25-26 = 0
k-3 = 0
k = 3

Let p(x) = x³-4x²-2x+20
By factor theorem, (x+2) is factor of p(x), if p(-2) = 0
p(-2) = (-2)³-4(-2)²-2(-2)+20   
{To find the zero of x+2; put x+2 = 0, we get x = -2}
= -8-4(4)+4+20
p(-2) = 0
Therefore, (x+2) is a factor of x³-4x²-2x+20

Let, p(x) = x³+13x²+ 32x + 20
Sum of all the coefficients = 1 + 13 + 32 + 20 = 66 ≠ 0
Hence, (x -1) is not a factor.
Sum of coefficients of even powers and constant term = 13 + 20 = 33
Sum of coefficients of odd powers = 1 + 32 = 33
Hence, (x +1) is a factor of p (x)
Now we use synthetic division to find the other factors

Let p(x) = 8x⁴- 2x² + 6x-7
Standard form = 8x⁴+ 0x³- 2x²+ 6x - 7
Co-efficients are 8 0 - 2 6 -7
Q(x) = 2x+1,
To find zero of 2x + 1, put 2x + 1 = 0; 2x = -1, x = \(-\frac{1}{2}\)
Synthetic division

Quotient = \(\frac{1}{2}\)[8x³-4x²+6] = 4x³-2x²+3
Quotient 4x³ - 2x² + 3 is compared with the given quotient 4x³-px²-qx+3
Co-efficients of x² is p = -2
Co-efficients of x is q = 0
Remainder is - 10

If 2x-3y−4z = 0 then 8x³− 27y³− 64z³=?
If x+y+z = 0 then x³+y³+z³ = 3xyz
8x³− 27y³− 64z³ = (2x)³+(−3y)³+(−4z)³
= 3\(\times\)2\(\times\)−3y\(\times\)−4z = 72xyz

Let P(x) = 2x³-9x²+x+12
By factor theorem,
(2x-3) is a factor of P(x), if \(P(\frac{3}{2})=0\)
\(P(\frac{3}{2})=2(\frac{3}{2})^3-9(\frac{3}{2})^3+\frac{3}{2}+12\)

ஃ (2x-3) is a factor of P(x) = 2x³-9x²+x+12

(i) (2x + 5), (5x + 2)
(2x + 5) = 1\(\times\)(2x+5)
(5x+2) = 1\(\times\)(5x+2)
ஃ G.C.D = 1
(ii) a^m+1, a^m+2,a^m+3
a^m+1 = a^m\(\times\)a¹
a^m+2 = a^m\(\times\)a¹\(\times\)a¹
a^m+3 = a^m\(\times\)a¹\(\times\)a¹\(\times\)a¹
G.C.D = a^m\(\times\)a¹
= a^m+1
(iii) 2a²+ a, 4a²-1
2a²+ a = a(2a + 1)
4a²- 1 = (2a)²-1² = (2a + 1)(2a - 1)
G.C.D = (2a + 1)
(iv) 3a²,5b³,7c⁴
3a² = 1\(\times\)a\(\times\)a
5b³ =1\(\times\)5b\(\times\)b\(\times\)b
7c⁴ =1\(\times\)7c\(\times\)c\(\times\)c\(\times\)c
ஃ G.C.D = 1
(v) x⁴-1, x²-1
x⁴-1 = (x²)²-1² = (x² + 1)(x²- 1)
= (x² + 1)(x²- 1²) = (x²+1)(x+1)(x-1)
x²-1 = x²-1² = (x+1)(x-1)
G.C.D = (x+1)(x-1) = x²-1
= x + 1
(vi) a³- 9ax², (a - 3x)²
a³-9ax² = a(a²-(3x)²) = a(a + 3x)(a - 3x)
(a - 3x)² = (a - 3x)(a - 3x)
G.C.D = (a - 3x)

R.H.S = \(\frac{1}{2}(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]\)
= \(\frac{1}{2}(x+y+z)[x^2-2xy+y^2+y^2-2yz+z^2-2zx+x^2]\)
= \(\frac{1}{2}[(x+y+z)[2x^2+2y^2+2z^2-2xy-2yz-2zx^2]]\)

= x³+y³+z³-3xyz = L.H.S
Hence it is verified.

We know that (x +a)(x +b)(x +c) = x³ + (a +b +c)x² + (ab +bc + ca) x +abc     .....(1)
(i) (x+5)(x+6)(x+4)
= x³ + (5 + 6 + 4) x²+ (30 + 24 + 20) x + (5)(6)(4)
= x² + 15x²+74x + 120
[Replace a by 5 b by 6 c by 4 in (1)]
(ii) (3x-1)(3x+2)(3x-4)
(3x)³+(-1+2-4)(3x)²+(-2-8+4)(3x)+(-1)(2)(-4)
= 27x³+(-3)9x²+(-6)(3x)+8
= 27x³- 27x²-18x + 8
[Repalce x by 3x, a by -1, b by 2, c by -4 in (1)]