(a+b+c)²=a²+b²+c²+2(ab+bc+ca)
(6)²=a²+b²+c²+2\(\times\)11
a²+b²+c²=36-22=14
a³+b³+c³-3abc=(a+b+c)[a²+b²+c²-(ab+bc+ca)]
= 6 \(\times\) (14-11)=6\(\times\)3=18.

Let each of the equal sides of the rhombus be a cm.
Then,      Perimeter = a + a + a + a = 4a m
According to the question,  4a = 200
\(\Rightarrow \)   a= \(\frac { 200 }{ 4 } \) = 50 m

d₁= 80 m
a²=\({ \left( \frac { { d }_{ 1 } }{ 2 } \right) }^{ 2 }+{ \left( \frac { { d }_{ 2 } }{ 2 } \right) }^{ 2 }\)
\(\Rightarrow \)   (50)² =( 40)²+\({ \left( \frac { { d }_{ 2 } }{ 2 } \right) }^{ 2 }\)
\(\Rightarrow \) \({ \left( \frac { { d }_{ 2 } }{ 2 } \right) }^{ 2 }\)= (50)^2-(40)² = 900= (30)²
\(\Rightarrow \)   \(\frac { { d }_{ 2 } }{ 2 } \)  = 30
\(\Rightarrow \)  d₂ = 60 m
\(\therefore \) Area of the rhombus = \(\frac { 1 }{ 2 } \) d₁d₂=\(\frac { 1 }{ 2 } \)\(\times \)80\(\times \)60 = 2400 m²

a:b:c = 5:12:13
5+12+13 = 30
a+b+c = 300 cm
\(\therefore a=\frac { 5 }{ 30 } \times 300=50\) cm
 \(b=\frac { 12 }{ 30 } \times 300=120\) cm
\(c=\frac { 13 }{ 30 } \times 300=130\)
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 50+120+130 }{ 2 } =150\) cm
\(\therefore \)  Area of the triangular field \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 150(150-50)(150-120)(150-130) } \)
\(\\ =\sqrt { 150\times 100\times 30\times 20 } \)
= 3000 cm²           ----(1)
Let the length of the perpendicular from the opposite vertex to the side whose length is 130 cm be h cm.Then, 
Area of the triangular field  \(=\frac { 130\times h }{ 2 } \) = 65h cm²   ...(2)
From (1) and (2)
65h = 3000
\(\Rightarrow h=\frac { 3000 }{ 65 } =\frac { 600 }{ 13 } \) cm = 46.15 cm

Let the third side be x cm. Then, 12 + 12 + x = 30
24 + x = 30
x = 6 cm
So, a = 12 cm, b = 12 cms, c = 6 cm
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 12+12+6 }{ 2 } \)
= 15 cm
\(\therefore \)   Area \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 15(15-12)(15-12)(15-6) } \)
 \(=9\sqrt { 15 } \) cm²

\((3x-5y-4)(9x^2+25y^2+15xy+12x-20y+16)\)
\(=\left\{ (3x)+(-5y)+(-4)(3x)^{ 2 }+(-5y)^{ 2 }+(-4)^{ 2 }-(3x)(-5y)-(-5y)(-4)-(-4)(3x) \right\} \)
\(=(3x)^2+(-5y)^3+(-4)^3-3(3x)(-5y)(-4)\)
\(=27x^3-125y^3-64-180xy\)

We have x=2y+6
\(\Rightarrow +(-2y)+(-6)=0\)
\(\therefore\)   \(x^3+(-2y)^3+(-6)^3=3x(-2y)(-6)\)
\(\Rightarrow x^3-8y^3-216=36xy\)
\(\therefore\)  \(x^3-8y^3-36xy-216=0\)

Let \(p(x)=x^{ 4 }+ax^{ 3 }-3x^{ 2 }+2x+b\)
If (x+1) and (x-1) are factors of p(x), then by factor theorem,
\(p(-1)=0 \ \ .......(1)\ x+1=0\ \ \ \Rightarrow \ x=-1\)
and   \(p(1)=0\ \ \ .......(2)|\ -1=0\Rightarrow x=1\)
Now, \(p(-1)=0\)
\(\Rightarrow (-1)^{ 4 }+a(-1)^{ 3 }-3(-1)^{ 2 }+2(-1)+b=0\)
\(\Rightarrow 1-a-3-2+b=0\)
\(\Rightarrow -a+b=4\ \ .......(3)\)
and   \(p(1)=0\)
\(\Rightarrow (1)^{ 4 }+a(1)^{ 3 }-3(1)^{ 2 }+2(1)+b=0\)
\(\Rightarrow 1+a-3+2+b=0\)
\(\Rightarrow a+b=0\ .......(4)\)
Solving (3) from (2), we get
a = -2,b = 2

\(\sqrt { 2 } (x+p)=0\) 
\(\Rightarrow x+p=0\) 
\(\Rightarrow x=-p\)
According to the question,
\(-p=-2\) 
\(\Rightarrow \ p=2\)
Let \(f(x)={ px }^{ 2 }+kx+2\sqrt { 2 } \)
Then, \(f(x)={ 2x }^{ 2 }+kx+2\sqrt { 2 } \)
If \(x=-2\) is a zero of f(x) then
\(f(-2)=0\) 
\(\Rightarrow \ 2{ (-2) }^{ 2 }+k(-2)+2\sqrt { 2 } =0\) 
\(\Rightarrow 2k=8+2\sqrt { 2 }\) 
\(\Rightarrow k=4+\sqrt { 2 } \)

\(100\sqrt { 3 } \)  cm²

\((3a-7b-c)^2\)
\((3a-7b-c)^{ 2 }={ \{ 3a+(-7b)+(-c)\} }^{ 2 }\)
\(={ (3a) }^{ 2 }+{ (-7b) }^{ 2 }+{ (-c) }^{ 2 }+2(3a){ (-7b) }+2(-7b)(-c)+2(-c)(3a)\)
\(=9{ a }^{ 2 }+49{ b }^{ 2 }+{ c }^{ 2 }-42ab+14bc-6ca\)

\(x^3-3x^2-9x-5\)
Let \(p(x)={ x }^{ 3 }-3{ x }^{ 2 }-9x-5\)
By trail, we find that
\(p(-1)={ (-1) }^{ 3 }-3{ (-1) }^{ 2 }-9(-1)-5\)
\(=-1-3+9-5=0\)
\(\therefore\) By Factor Theorem, x-(-x), i.e., (x+1) is a factor of p(x).
Now,
\({ x }^{ 3 }-3{ x }^{ 2 }-9x-5={ x }^{ 2 }(x+1)-4x(x+1)-5(x+1)\)
\(=(x+1)({ x }^{ 2 }-4x-5)\)
\(=(x+1)({ x }^{ 2 }-5x+x-5)\)
\(=(x+1)\{ x(x-5)+1(x-2)\} \)
\(=(x+1)(x-5)(x+1).\)

Let \(p(x)=x^3-ax^2+6x-a\)
\(x-a=0\)
\(\Rightarrow x=a\)
\(\therefore\)  Remainder = \((a)^3-a(a)^2+6(a)-a\)          
=  \(a^3-a^2+6a-a\)
= 5a

\(x+\pi\)
\(x+\pi=0\ d\Rightarrow\ x=-\pi\)
\(\therefore\) Remainder \(= (-\pi)^3+3(-\pi)^2+3(-\pi)+1\)
 \(= -\pi^3+3\pi^2+3\pi+1\)

\(x\)
\(\therefore\) Remainder
 \(= (0)^3+3(0)^2+3(0)+1=1\)