\((p+q+r)^2+(p-q-r)^2\)
\(=(p+q+r)^{ 2 }+{ \{ p+(-q)+(-r)\} }^{ 2 }\)
Using Identify V
\(=(p^{ 2 }+q^{ 2 }+r^{ 2 }+2pq+2qr+2rp)+( p^{ 2 }+q^2+r^2-2pq+2qr-2rp)\)
\(=(p^2+q^2+r^2+2qr)\)

\(2(a^2+b^2)=(a+b)^2\)
\(\Rightarrow 2a^2+2b^2=a^2+b^2+2ab\)
Using Identify
\(\Rightarrow a^2+b^2-2ab=0\)
\(\Rightarrow (a-b)^2=0\)
\(\Rightarrow\ a-b=0\)
\(\Rightarrow a=b\) 

\(a(x-a)=2ab-b(x-b)\)
\(\Rightarrow ax- a^2=2ab-bx+b^2\)
\(\Rightarrow ax+bx=a^2+2ab+b^2\)
\(\Rightarrow x(a+b)=(a+b)^2\) Using Identify I
\(x=a+b\)
Cancelling (a+b) from both sides as (a+b) ≠ 0
Hence the given equation has a solution x=a+b.

We have
\((-11)+(8)+(3)=0\)
Therefore, 
\((-11)^3+(8)^3+(3)^3\)\(=3(-11)(8)(3)\)
=-792   | From Identify VII, if x+y+z=0
Then  \(x^3+y^3+z^3=3xyz\)

105 x 93
=(100+5) x (100 x7)
=(100+5) x {100+(-7)}
=(100)²+{5+(-7)}(100)+(5)(-7) | Using Identity IV
=10000-200-35=9765

\(p(x)=2{ x }^{ 3 }-3{ x }^{ 2 }+ax-3a+9\)
By remainder theorem,
\(p(-1)=16\ x+1=0\quad \Rightarrow x=-1\)
\(\Rightarrow 2{ (-1) }^{ 3 }-3{ (-1) }^{ 2 }+a(-1)-3a+9=16\) 
\(\Rightarrow -2-3-a-3a+9=16\)
\(\Rightarrow 4a=-12\) 
\(\Rightarrow a=-3\) 
\(\therefore p(x)=2{ x }^{ 3 }-3{ x }^{ 2 }-3x-3 \times(-3)+9\) 
\(={ 2x }^{ 3 }-3{ x }^{ 2 }-3x+18\)
\(\therefore \) Remainder when p(x) is divided by x+2 =p(-2)
By remainder theorem: \(x+2=0\Rightarrow x=-2\)
\(={ 2(-2) }^{ 3 }-3{ (-2) }^{ 2 }-3(-2)+18\) 
\(=-16-12+6+18=-4\) 

Divisor = x + 3
\(x+3=0\ \Rightarrow x=-3\)
\(\therefore\)Remainder= p(-3) | By remainder theorem
\(= k{ (-3) }^{ 3 }+9{ (-3) }^{ 2 }+4(-3)-8\) 
\(=-27k+81-12-8\) 
\(=-27k+61\)
According to the question,
\(-27k+61=10(1-k)\) 
\(\Rightarrow -27k+61=10-10k\) 
\(\Rightarrow 17k=51\) 
\(\Rightarrow k=3\) 

\(\sqrt { 2 } (x+p)=0\) 
\(\Rightarrow x+p=0\) 
\(\Rightarrow x=-p\)
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
\(f(-2)=0\) 
\(\Rightarrow \ 2{ (-2) }^{ 2 }+k(-2)+2\sqrt { 2 } =0\) 
\(\Rightarrow 2k=8+2\sqrt { 2 }\) 
\(\Rightarrow k=4+\sqrt { 2 } \)

Number of terms = 3
Terms: 3x², -5x, 7

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