(a) ∵ Area of square field = Side \(\times\) Side =(Side)²
= x² - 26x + 169
So, we factorise x² - 26x + 169
=x²-2\(\times\)13\(\times\)x+(13)²
=(x-13)²
So, area of square field =(Side)² =(x-13)²
Side=(x-13)unit
(b) If x =15, then, side =15-13 = 2 unit
So, perimeter of square field = 4 \(\times\) 2 = 8 units
[∵ Perimeter of a square = 4 \(\times\) side]
(c) He believes in equality and does not differentiate between a girl child and a boy child.

It is given that, the curved surface area of the cylinder = 2\(\pi \)(y²+ 7y + 12).....(i)
We know that, the formula of curved surface area of the cylinder = 2\(\pi \)rh.....(ii)
where, r = Radius of the cylinder
h = Height of the cylinder
On comparing Eqs. (l) and (ii), we get
2\(\pi \)rh = 2\(\pi \)(y²+ 7y + 12)
r\(\times\)h=(y²+7y+12).....(iii)
But radius of the cylinder = R = (y + 3)
Putting this value in Eq.(iii), we get
(y+ 3)\(\times\)h =(y² + 7y +12)
\(h=\frac { \left( { y }^{ 2 }+7y+12 \right) }{ \left( y+3 \right) } \)
We have to factorise (y²+ 7y + 12)
Since, 4 \(\times\) 3 = 12 and 4 + 3 = 7
Putting this value in (y² + 7y + 12), we get
y² + (4 + 3)y+ 4 \(\times\) 3
= y²+4y+3y+4x³
= y(y + 4)+ 3(y + 4)=(y + 4)(y + 3)
\(h=\frac { \left( { y }^{ 2 }+7y+12 \right) }{ \left( y+3 \right) } =\frac { \left( y+4 \right) \left( y+3 \right) }{ \left( y+3 \right) } \)
= (y + 4) units
Hence, height of the cylinder = (y + 4) units.

p⁴+ q⁴+p²q²
= p² \(\times\) p² + q² \(\times\) q² + p²q²
= (p²)² + (q²)² + p²q²
= (p²)² + (q²)² + p²q² + p²q²- p²q²
[adding and subtracting p²q²]
= (p²)² + (q²)² + 2 \(\times\) p² \(\times\) q²-p²q²
=(p² + q²)² -p²q² [using identity, (a² + b² + 2ab) = (a + b)² where, a = p², b = q²
= (p² + q²)²-(pq)²
= (p² + q² + pq)(p² + q² -pq) [by using identity, (a² - b²) =(a + b)(a-b) where, a=(p²+q²),b=pq]

It is given that,
Product of the two expressions = x⁵ + x³ + x
One of them = x² + x + 1
\(\therefore other\ expression=\frac { Product\ of\ two\ expressions }{ one\ expressions } \)
\(=\frac { { x }^{ 5 }+{ x }^{ 3 }+x }{ { x }^{ 2 }+x+1 } \)
Now, we have to factorise x⁵ + x³ + x
= x (x⁴+x²+1)
= x {(x²)² + (1)² + x²}
= x {(x²)² + (1)²+ x² + x² -x²}
[adding and subtracting x²]
= x {(x²)² + (1)² + 2(x²)(1) -x²}
= x {(x² + 1)² -(x)²}
[by using identity, (a + b)² = a² + b² + 2ab where, a = x², b =1]
= x (x² + 1 + x)(x² + 1 - x) [by using identity]
(a² - b²) =(a + b)(a -b), where, a = x² +1,b=x]
\(\frac { { x }^{ 5 }+{ x }^{ 3 }+x }{ { x }^{ 2 }+x+1 } =\frac { x\left( { x }^{ 2 }+x+1 \right) \left( { x }^{ 2 }-x+1 \right) }{ \left( { x }^{ 2 }+x+1 \right) } \)
= x(x² - x++)

(i) (ab + bc) (ab - bc) + (bc + ca) (be - ca)+(ca+ab)(ca-ab)=0
LHS = (ab + bc)(ab - bc) + (bc + ca)(bc - ca)+ (ca + ab)(ca - ab)
= {(ab)² -(bc)²} + {(bc)² -(ca)²} + {(ca)² - (ab)²}
[by using identity, (A² - B²) = (A + B)(A - B)]
= a²b² -b²c² + b²c²-c²a² + c²a²-a²b²
= a²b² -a²b² + b²c² -b²c² + c²a² -c²a²
= 0= RHS
LHS = RHS
(ii) (a + b +c)(a² + b² + c² -ab-bc -ca)
= a³ + b³ + c³ - 3abc
LHS
= (a + b + c) \(\times\) (a² + b² + c²- ab - bc - ca)
= a³ + ab² + ac² -a²b - abc - ca²+ ba² + b³ + bc²- ab² - b²c - abc+ ca² + b²c + c³ - abc - bc² - ac²
= a³ + b³ + c³-abc - abc - abc + ab²- ab²+ ac² -ac² + a²b -a²b + ca² -ca²+ bc² - bc² + b²c - b²c
= a³ + b³ + c³- 3abc = RHS
LHS = RHS

(m+n)(m-n)(m²+n²)

(p-3)(p²+2-q)

Area of square x⁴=18x² + 81 sq unit 
Area if x = 2 is 25 sq units.

Number of observations = 4x²+9 

p=4

4

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