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 8p³+ q³
This can be written as = 23p³+ q³
= (2p)³+ q³
Comparing this with a³+ b^3, we get a = 2p,b = q
We know a³+ b³ = (a+b) (a²−ab+b²)
(2p)³+ q³ = (2p+q)[(2p)²−(2p)(q)+q²]
8p³+ q³ = (2p+q) [4p²−2pq+q²]

This is in the form of ax² + bx + c
We get a = 7,b = 2,c = −5
Now, the product = a x c = 7 x (−5) = –35 and sum b = 2
= 7c² + 2c − 5
= (7c² − 5c) + (7c − 5) (the middle term 2c can be written as – 5c + 7c)
= c(7c − 5) + 1(7c−5) (taking out the common factor 7c – 5 )
= (7c − 5)(c + 1)
Therefore, (7c–5), (c+1) are the two factors.

Now, 49x²- 64y² = 72x² - 8²y²
= (7x)² - (8y)²
Comparing this with a²-b² = (a + b)(a - b)
we get a = 7x, b = 8y
(7x)²- (8y)² = (7x + 8y)(7x - 8y)

Now, x² + 8x + 16
This can be written as x² + 8x + 42
Comparing this with a² + 2ab + b² = (a + b)² we get a = x; b = 4
(x²) + 2(x)(4) + (4)² = (x + 4)²
x² + 8x + 16 = (x + 4)²

We have (2x + 5)(x - y) + (4y)(x-y)
Taking out the common binomial factor (x-y)
We get,(x - y)(2x + 5 + 4y)

We have,2m3-5m2+9m taking out the common factor ‘m’ from each term, we get
= 3(2m2-5m+9)

\(speed=\cfrac { distancetravelled }{ timetaken } \)
= \(\cfrac { { 4x }^{ 2 }+3{ xy }^{ 2 }+5x }{ 2x } \)

= \({ 2x }^{ 2-1 }+\cfrac { 3 }{ 2 } { y }^{ 2 }+\cfrac { 5 }{ 2 } \)
Speed = \(\left( 2x+\cfrac { 3 }{ 2 } { y }^{ 2 }+\cfrac { 5 }{ 2 } \right) \) km/hr

Total number of pictures = 100x²y³
Each page contains = 4xy pictures

= 25xy² pages

Mark the unshaded parts of the given figure as I, II, III and IV
Area of the I and III parts = Area of the square – Area of 2 semicircles
\(={ a }^{ 2 }-\left( 2\times \frac { 1 }{ 2 } { \pi r }^{ 2 } \right) \)
\(=(10\times 10)-\left( \frac { 22 }{ 7 } \times 5\times 5 \right) \) = 100 - 78.57 = 21.43 cm²
Similarly, the area of the II and IV parts = 21.43 cm²
∴ Area of the unshaded parts (I, II, III and IV)
= 21.43 x 2 = 42.86 cm² 
∴ Area of the shaded part = area of the square – area of the unshaded parts
= 100 – 42.86 = 57.14 cm² 

This is one of the solutions. Try for more

This is one of the solutions. Try for more.

Figures I and II are trapeziums separately as well as combinedly.
The parallel sides of the combined trapezium (I and II) are 5 cm and 16 cm. Its height, h = 8 + 8 = 16cm
Length of the rectangle = 16 cm
Breadth of the rectangle = 8 cm
∴  Area of the combined invitation card
= area of the combined trapezium + area of the rectangle
\(=\left( \frac { 1 }{ 2 } h\times (a+b) \right) +(l\times b)\)
\(\left( \frac { 1 }{ 2 } \times 16\times (5+16) \right) +(16\times 8)\)
= 168 + 128 = 296cm²
Aliter:
Area of the invitation card = area of the outer rectangle – area of the right angled triangle
\(= l\times b - \frac { 1 }{ 2 } \times h \times b \)
= \(24 \times16 \frac{1}{ 2} \times11 \times16\)
= − 384 88 296 = cm²

Side of the square = 5 cm
Diameter of the semi-circle = 5 cm
∴ Radius = 2.5 cm
Side of the equilateral triangle = 5 cm
∴  Area of the keychain = area of the semi circle + area of the square + area of the equilateral triangle
\(=\frac { 1 }{ 2 } { \pi r }^{ 2 }+{ a }^{ 2 }+\frac { \sqrt { 3 } }{ 4 } { a }^{ 2 }\)
\(=\left( \frac { 1 }{ 2 } \times 3.14\times 2.5\times 2.5 \right) +\left( 5\times 5 \right) +\left( \frac { \sqrt { 3 } }{ 4 } \times 5\times 6 \right) \)
= 9.81 + 25 + 10.83
= 45.64cm² (approx.)

(i) Area of the blue shaded part = Area of the quadrant of a circle
\(=\frac { 1 }{ 4 } \times { \pi r }^{ 2 }\)
\(=\frac { 1 }{ 4 } \times 3.14\times 2\times 2\)
= 3.14 cm² (approximately)
(ii) Area of the grey shaded part =  Area of the square –  Area of the blue shaded part
a² - \(\frac{1}{4}\pi r^2\)
= 6 x 6 - 3.14
= 36 - 3.14
= 32.86 cm² (approximately)