\(\cfrac { -21abc^{ 2 } }{ 7abc } =\cfrac { -21 }{ 7 } { a }^{ 1-1 }{ b }^{ 1-1 }{ c }^{ 2-1 }=-3{ a }^{ 0 }{ b }^{ 0 }{ c }^{ 1 }=-3{ { a }^{ 0 }{ b }^{ 0 }{ c }^{ 1 } }=-3c\)

Comparing this with (x + a)(x + b)(x + c), we get x = x, a = 3 ,b = 5 , c = 2
(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc
(x + 3)(x + 5)(x + 2) = (x)³ + (3 + 5 + 2)(x)² + (3 × 5 + 5 × 2 + 2 × 3)x + (3)(5)(2)
= x³ + 10x²+(15 +10 + 6)x + 30
(x + 3)(x + 5)(x + 2) = x³ + 10x² + 31x + 30

Comparing it with (a + b)(a - b) we get a = 3x b = 5y
Now (a + b)(a - b) = a²-b²
(3x + 5y)(3x - 5y) = (3x)² - (5y)² (replacing a and b values)
= 3² x² - 5² y²
(3x + 5y) (3x - 5y) = 9x² - 25y²

We have, (2x²y²) × (3y²z) × (−z²x3)
= (+) × (+) × (−)(2 × 3 × 1)(x² × x³)(y² × y²)(z × z²)
=  − 6x⁵y⁴z³

3y(x−2)²−2(2− x)
= 3y (x - 2) (x - 2) + 2 (x - 2) = (x -2) [3y (x -2) + 2]

(104)³ = (100 + 4)³
Comparing (100 + 4)³ with (a + b)³ we have a = 100 and b = 4
(a + b)³ = a³ + 3a²- b + 3ab² + b³
(100 + 4)³ = (100)³ + 3 (100)² (4) + 3 (100) (4)² + 4³
= 10,00,000 + 3(10000) 4 + 300(16) + 64
= 10,00,000 + 1,20,000 + 4,800 + 64 = 11,24,864

6xy + 3xy = 9x²y² (false)
6xy + 3xy = 9 xy

\(\cfrac { { x }^{ 3 }{ y }^{ 2 } }{ { x }^{ 2 }y } ={ x }^{ 3-2 }{ y }^{ 2-1 }={ x }^{ 1 }{ y }^{ 1 }=xy\)

b x (−15m²n³p) = 45m³n³p²
\(\mathrm{b}=\frac{45 m^{3} n^{3} p^{2}}{-15 m^{2} n^{3} p}\)
b = -3mp

(2x + 3) (2x - 4) 
= 4x² - 8x + 6x - 12 
= 4x² - 2x - 12

-10p³+6p²-14p

I x b x h = (4pq²) x (^_3p² q) x (2p³q³)
= -24p⁶q⁶

a³ \(\times\) (-4² b) = (-4) \(\times\) (a³ \(\times\) a²) \(\times\) (b) = -4a⁵b

6x \(\times\) 4= (6 \(\times\) 4)(x) = 24x

We have 81x³-3y³ = 3(27x³-y³)
= 3(3³x³-y³)
= 3[(3x)³-y³]
Comparing this a³-b³,we get a = 3x, b = y
= 3{(3x-y)[(3x)²+(3x)(y)+y²]}
Therefore, 81x³-3y³ = 3{(3x-y)[9x²+3xy+y²]}

We have, x² + yz + xy + xz
Group the terms suitably as = (x² + xy) + (yz + xz)
= x(x + y) + z(y + x) (addition is commutative)
= x(x + y) + z(x + y) [taking out the common factor(x + y)]