\((a+b)^{2}=a^{2}+b^{2} +2 a b \)
\(\left(x+\frac{1}{x}\right)^{2} =x^{2}+\frac{1}{x^{2}}+2 x \frac{1}{x} \)
 = 23 + 2 = 25 
\( \therefore x+\frac{1}{x} =\pm 5\) 
\( \left(x+\frac{1}{x}\right)^{3} =x^{3}+\frac{1}{x^{3}}+3(x)\left(\frac{1}{x}\right)(x+\frac{1}{x}) \)
\(5^{3} =x^{3}+\frac{1}{x^{3}}+3\left(x+\frac{1}{x}\right) \)
\(125 =x^{3}+\frac{1}{x^{3}}+3(5) \)
\(125-15 =x^{3}+\frac{1}{x^{3}} \)
\(x^{3}+\frac{1}{x^{3}} =\pm 110\)

\( (x+y)^{3}=x^{3}+y^{3}+3 x y(x+y) \)
\(a^{3}+\frac{1}{a^{3}}+3(a)\left(\frac{1}{a}\right)\left(a+\frac{1}{a}\right)=\left(a+\frac{1}{a}\right)^{3} \)
\(a^{3}+\frac{1}{a^{3}}+3(6) =6^{3} \)
\(a^{3}+\frac{1}{a^{3}}+18 =216 \)
\(a^{3}+\frac{1}{a^{3}}=216-18 =198 \)
\(\therefore a^{3}+\frac{1}{a^{3}} =198\)

We know that (a+b)³ ≡ a³+3a²b+3ab²+b³
(2a+3b)³ = (2a)³+3(2a)²(3b)+3(2a)(3b)²+(3b)³
= 8a³+36a²b+54ab²+27b³

(4m+5)(4m+4)(4m-5) = (4m)³+(5+4-5)(4m)²+[(5x4)+(4x-5)+(-5x5)](4m)+(5\(\times\)4\(\times\)-5)
= 64m³+64m²+(20-20-25)4m-100
= 64m³+64m²-100m-100

(a+b+c)²= a²+ b²+ c²+ 2ab + 2bc + 2ca
ஃ (x + 2y + 3z)²= (x)²+(2y)²+(3z)²+2(x)(2y)+2(2y)(3z)+2(3z)(x)
= x² + 4y² + 9z² + 4xy + 12yz + 6xz

We know that, if a + b + c = 0, then a³+b³+ c³ = 3abc
Here, a+b+c =10 -15 + 5 = 0
Therefore, 10³+(-15)³+5³ = 3(10)-(15)(5)   [Replace a by 10, b by -15, c by 5]
10³-15³+5³ = -2250

We know that, (a+b+c)(a²+b²+c²-ab-bc-ca) = a³+ b³+ c³-3abc
(2x+3y+4z)(4x²+9y²+16z²-6xy-12yz-8zx) = (2x)³+(3y)³+(4z)³-3(2x)(3y)(4z)
= 8x³+27y³+64z³- 72xyz

We know that,
(a+b+c)² = a²+b²+c²+ 2ab + 2bc + 2ca
Substituting, a = 2x, b = 3y and c = 4z
(2x+3y+4z)² = (2x)²+(3y²)+(4z)²+2(2x)(3y)+2(3y)(4z)+2(4z)(2x)
= 4x²+ 9y²+16z²+12xy + 24yz +16xz

Replacing ‘b’ by ‘-b ’ in the expansion
(a+b+c)² = a²+ b²+ c²+ 2ab + 2bc + 2ca
(a+(-b)+c)² = a²+(-b)²+ c²+ 2a(-b) + 2(-b)c+ 2ca
= a²+ b²- 2ab - 2bc + 2ca

(5x + 4y)(5x −4y)    [we have (a+b)(a-b) = a²-b²]
(5x+4y)(5x-4y) = (5x)²-(4y)²    put [a = 2a, b = 4y]
= 25x²-16y²