We have 50x² - 98 = 2(25x² - 49)
= 2[(5x)² - (7)²] == 2[(5x + 7)(5x - 7)] [Using a² - b² = (a + b)(a - b)]
= \(\frac { 81x^{ 3 }[50x^{ 2 }-98] }{ 27x^{ 2 }[5x+7] } \)
=\(\frac { { 81x }^{ 3 } }{ { 27x }^{ 2 } } \left[ \frac { 2(5x-7)(5x+7) }{ (5x+7) } \right] =3x[2(5x-7)]\)
= 3x\(\times \)2\(\times \)(5x- 7) = 6 x (5x - 7)

We have 27x³ = 3 x 3 x 3 \(\times \)x\(\times \)x\(\times \)x
21x² = 3\(\times \)7\(\times \)x\(\times \)x
15x⁴ = 3 x 5 \(\times \) x \(\times \) x \(\times \) x \(\times \) x
We get common factors as 3, x and x.
i.e., 3 \(\times \) x \(\times \)x or 3x²
∴27x³ =: 3x² x 3 \(\times \) x \(\times \) 3 = 3x² \(\times \)9x
21x²=\({ 3x }^{ 2 }\times 7={ 3x }^{ 2 }\times 7\)
\({ 15x }^{ 4 }=3x^{ 2 }\times 5x\times x={ 3x }^{ 2 }\times { 5x }^{ 2 }\)
\(\Rightarrow { 27x }^{ 3 }-{ 21x }^{ 2 }+{ 15x }^{ 4 }\)
\(\Rightarrow { (3x }^{ 2 }\times 9x)-({ 3x }^{ 2 }\times 7)+{ (3x }^{ 2 }\times { 5x }^{ 2 })\)
\({ 3x }^{ 2 }[9x-7+{ 5x }^{ 2 }]\)
 

∵ z²+ 6z - 16 = z²+ 8z - 2z - 16
∵ 8 - 2 = 6
= z(z + 8) - 2(z + 8)8 x 2 = 16
= (z + 8)(z - 2)
\(\therefore \frac { 4yx({ z }^{ 2 }+6z-16) }{ 2y(z+8) } \)
\(=\frac { 2\times 2\times y\times z\times (z+8)\times (z-2) }{ 2\times y\times (z+8) } \)
\(=\frac { 2\times z\times (z-2) }{ 1 } =2z(z-2)\)
thus, 4yz(z2 + 6z-16)\(\div \)2y(z + 8) = 2z(z - 2)

RHS = (9982)² - (18)²
= (9982 + 18)(9982 -18)
[using identity a² - b² = (a + p)(a -b)]
= 10000 x 9964
LHS = 10000 x x
Since, it is given that
LHS = RHS
10000 x x =10000 x 9964
\(\Rightarrow\) \(x={10000 \times 9964 \over 10000}=9964\)

9xy(x⁴+y⁴) sq units

Number of chocolates Abhinav bought = (x + 4)
Cost of one chocolate = Rs.(x + 4)
Total amount paid by him = Number of chocolates \(\times\) Cost of one chocolate
= Rs. ( x + 4) x (x + 4) = Rs.(x + 4)²
= Rs.{(x² + (4)² + 2 \(\times\) x \(\times\) 4))
[using identity, (a + b)² = (a² + b² + 2ab)]
= Rs.(x² + 4\(\times\)4+ 8x)
= Rs.(x² + 16 + 8x)
It is given that, if x = 10 then,
the amount paid by him
= Rs. {(10)² + 16 + 8 \(\times\)10}
= Rs.(10 x10 + 16+ 8 \(\times\)10)
= Rs.(100+16+ 80)=Rs.196
(ii) The value depicted here is that Abhinav is helpful and friendly by nature to his friend.

3\(\pi \) unit 

If we compare the R.H.S. of Identity (IV) with x2 + 5x + 6, we find ab = 6, and a + b = 5. From this, we must obtain a and b. The factors then will be (x + a) and (x + b).
If a b = 6, it means that a and b are factors of 6. Let us try a = 6, b = 1. For these values a + b = 7, and not 5, So this choice is not right.
Let us try a = 2, b = 3. For this a + b = 5 exactly as required.
The factorised form of this given expression is then (x +2) (x + 3).

We have 3z - 24 = 3(z - 8)
∴ 4x²y²(3z - 24) \(\div \) 36xy(z - 8)
= \(\frac { 4{ x }^{ 2 }y^{ 2 }(3z-24) }{ 36xy(z-8) }\) 
\(=\frac { { 4x }^{ 2 }y^{ 2 }(3)(z-8) }{ 4xy\times 9(z-8) }\) 
\(=\frac { xy }{ 3 } \)

50y² - 98 = 2(25y² - 49)
=  2[(5y)² - (7)²]
=2[(5y - 7)(5y + 7)]
\(\therefore \ \frac{39 y^{3}\left(50 y^{2}-98\right)}{26 y^{2}(5 y+7)}=\frac{3 \times 13 \times y \times y \times y \times 2 \times(5 y-7) \times(5 y+7)}{2 \times 13 \times y \times y \times(5 y+7)} \)
\(=\frac{3 \times y \times(5 y-7)}{1}\)
Thus, 39y³(50y² - 98) / 26y²(5y + 7) = 3y(5y - 7)

m(m +1) (m - 1)

(x -5)²

Given mathematical statement is incorrect.
Hence, correct statement is
5y + 2y + y - 7y = (5 + 2 + 1)y - 7y = 8y - 7y = y
because coefficient 1 of a term is usually not shown. But while adding like terms, we include it in the sum.

12

ab - a - b + 1
= a(b - 1) - 1(b - 1)
= (a - 1)(b - 1).

10x² - 18x³ + 14x² = 2x² (5 - 9x + 7x²).

3x + 15z = 3x + 3  x 5z = 3 (x + 5z)
By using distrubutibe property ab +ac = a (b+c)]

105² - 103² = (105 +103)(105-103)
=208 x 2  = 416

We know what
Volume of a rectangular box or volume of a cuboid
= length x breadth x Height
= 2x x 3y x 4z = 24xyz