(998)³
(998)³=(1000-2)³
= (1000)³-(2)²-3(1000)(2)(1000-2) | Using Identity VII
= 1000000000-8-6000(1000-2)
= 1000000000-8-6000000+12000+
​​​​​​​= 994011662

\({ \left[ \frac { 3 }{ 2 } x+1 \right] }^{ 3 }\)
\(={ \left( \frac { 3 }{ 2 } x \right) }^{ 3 }+{ (1) }^{ 3 }+3\left( \frac { 3 }{ 2 } x \right) (1)\left( \frac { 3 }{ 2 } x+1 \right) \) | Using Identity VI
\(=\frac { 27 }{ 8 } { x }^{ 3 }+1+\frac { 9 }{ 2 } x\left( \frac { 3 }{ 2 } x+1 \right) \)
\(=\frac { 27 }{ 8 } { x }^{ 3 }+1+\frac { 27 }{ 4 } { x }^{ 2 }+\frac { 9 }{ 2 } x\)
\(=\frac { 27 }{ 8 } { x }^{ 3 }+\frac { 27 }{ 4 } { x }^{ 2 }+\frac { 9 }{ 2 } x+1\)

\((-2x+5y-3z)^2\)
\((-2x+5y-3z)^{ 2 }={ \{ (-2x)+5y+(-3z)\} }^{ 2 }\)
\(={ (-2x) }^{ 2 }+{ (5y) }^{ 2 }+{ (-3z) }^{ 2 }+2(-2x)(5y)+2(5y)(-3z)+2(-3z)(-2x)\)
\(=4{ x }^{ 2 }+25{ y }^{ 2 }+9{ z }^{ 2 }-20xy-30yz+12zx\)

\(\therefore \ p(0)={ (0) }^{ 2 }-(0)+1=1\)
\( p(1)={ (1) }^{ 2 }-(1)+1=1\)
and \(p(2)={ (2) }^{ 2 }-(2)+1=4-2+1=3\)

Term with the highest power of x = 5x³
Exponent of x in this term           = 3
Therefore Degree of this polynomial = 3

We have 4x² + y² + z² – 4xy – 2yz + 4xz = (2x)² + (–y)² + (z)² + 2(2x)(–y) + 2(–y)(z) + 2(2x)(z)
= [2x + (–y) + z]² (Using Identity V)
= (2x – y + z)² = (2x – y + z)(2x – y + z)
So far, we have dealt with identities involving second degree terms. Now let us extend Identity I to compute (x + y)³. We have:
(x + y)³ = (x + y) (x + y)²
= (x + y)(x² + 2xy + y2)
= x(x² + 2xy + y²) + y(x² + 2xy + y²)
= x³ + 2x²y + xy² + x²y + 2xy² + y³
= x³ + 3x²y + 3xy² + y3
= x³ + y³ + 3xy(x + y)

Using Identity V, we have
(4a – 2b – 3c)² = [4a + (–2b) + (–3c)]²
= (4a)² + (–2b)² + (–3c)² + 2(4a)(–2b) + 2(–2b)(–3c) + 2(–3c)(4a)
= 16a² + 4b² + 9c² – 16ab + 12bc – 24ac

As x – 1 is a factor of p(x) = 4x³ + 3x² – 4x + k, p(1) = 0
Now, p(1) = 4(1)³ + 3(1)² – 4(1) + k
So, 4 + 3 – 4 + k = 0
i.e., k = –3

Let p(x) = x² – 2x
Then p(2) = 2² – 4 = 4 – 4 = 0
and p(0) = 0 – 0 = 0
Hence, 2 and 0 are both zeroes of the polynomial x² – 2x.

\(\left(-\frac{1}{2}\right)\)

RHS = (x-y) (x²+y²+xy)
=x³+xy²+x²y-x²y-y³-xy²
= x³-y³= LHS
Now, 216x³-125y³=(6x)³-(5y)³
= (6x-5y)[(6x)²+(5y)²+6x \(\times\) 5y]
=(6x-5y)(36x²+25y²+30xy).

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  \(x^3+y^3+z^3=3xyz\)

Let \(p(x)={ x }^{ 3 }-6{ x }^{ 2 }+11x-6\)
By trial, we find that
\(p(1)={ (1) }^{ 3 }-6{ (1) }^{ 2 }+11(1)-6=0\)
\(\therefore\) By converse of factor theorem, (x-1) is a factor of p(x).
Now,  \({ x }^{ 3 }-6{ x }^{ 2 }+11x-6\)
\(={ x }^{ 2 }(x-1)-5x(x-1)+6(x-1)\)
\(=(x-1)({ x }^{ 2 }-5x+6)\)
\(=(x-1)({ x }^{ 2 }-2x-3x+6)\)
\(=(x-1)\{ x(x-2)-3(x-2)\} \)
\(=(x-1)(x-2)(x-3)\) 

\(4a^2+9b^2+c^2+12ab+4ac+6bc\)
\(=(2a)^2+(3b)^2+(c)^2+2(2a)(3b)+2(3c)(c)+2(c)(2a)\)
\((2a+3b+c)^2\)

525²-475² 
=(525+475)(525-475)=50000

\(f(x)=x^3+3x^2+bx-6\)
\(f(3)=0\)
\(\Rightarrow \ 3^3-3 \times3^2+b\times 3-6=0\)
\(\Rightarrow\ b=2\)

\(cx+d=0\ \Rightarrow\ x=-\frac{d}{c}\)

The terms containing x² is-x² , i.e., (-1) x²

A polynomial of degree n has maximum number of terms as (n+1)

Definition of cubic polynomial

(i) 2 + x² + x
The co-efficient of x² is 1.
(ii) 2 - x² + x3
The co-efficient of x² is (-1).
⁽ⁱⁱⁱ⁾ \(\frac{\pi}{2}\)x²+ x
The co-efficient of x² is \(\frac{\pi}{2}\).
(iv) \(\sqrt{2}\)x-1
∵  \(\sqrt{2}\)x-1 ⇒  \(\sqrt{2}\)x-1 + 0・x²
∴ The co-efficient of x² is 0.

0

\(-\frac { 31 }{ 4 } \)