\(a^2+b^2-2(ab-ac+bc)\)
\(=a^2+b^2-2ab+2ac-2bc\)
\(=(a-b)^2+2c(a-b)\)
\((a-b)(a-b+2c).\)

(999)³
=(1000-1)³
= (1000)³-(1)³-3(1000)(1)(1000-1)
= 1000000000-1-3000 X 999
= 1000000000-1-2997000
= 997002999

We know that
\((x+2y)^3=(x)^3+(2y)^3+3(x)(2y)(x+2y)\) Using  Identity VI
\(\Rightarrow(x+2y)^3=x^3+8y^3+6xy(x+2y)\)
\(\Rightarrow (10)^3=x^3+8y^3+6(15)(10)\)
\(\Rightarrow 1000=x^3+8y^3+900\)
\(\Rightarrow x^3+8y^3=1000-900=100\) 

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

\(p(z)={ 3z }^{ 2 }=4z+\sqrt { 17 } \)
\(\therefore p(3)=3{ (3) }^{ 2 }-4(3)+\sqrt { 17 }\) 
\(=15+\sqrt { 17 }\)

Number of terms = 4
Terms: t, -7t², 5, -t³

1, -3, 4

\(\left( { y }^{ 2 }+\frac { 3 }{ 2 } \right) \left( { y }^{ 2 }-\frac { 3 }{ 2 } \right) ​​\)
\(\left( { y }^{ 2 }+\frac { 3 }{ 2 } \right) \left( { y }^{ 2 }-\frac { 3 }{ 2 } \right) ​​=\left( z+\frac { 3 }{ 2 } \right) \left( z-\frac { 3 }{ 2 } \right) ​​\)| Where \(y^2=z\)
\(={ (z) }^{ 2 }-{ \left( \frac { 3 }{ 2 } \right) }^{ 2 }\) | Using identity III
\(={ z }^{ 2 }-\frac { 9 }{ 4 } ={ ({ y }^{ 2 }) }^{ 2 }-\frac { 9 }{ 4 } \) | Substituting the value of z
\(={ y }^{ 4 }-\frac { 9 }{ 4 } .\)

\(12x^2-7x+1\)
\(12x^{ 2 }-7x+1=12x^{ 2 }-4x-3x+1\)
\(=4x(3x-1)-1(3x-1)\)
\(=(3x-1)(4x-1)\)

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