\(\sqrt [ 3 ]{ 126 } ={ (125) }^{ 1/3 }=(125+1)^{ 1/3 }=\left\{ 125\left( 1+\frac { 1 }{ 125 } \right) \right\} ^{ 1/3 }=(125)^{ 1/3 }\left[ 1+\frac { 1 }{ 125 } \right] ^{ 1/3 }\)
\(=5\left[ 1+\frac { 1 }{ 3 } \times \frac { 1 }{ 125 } +... \right] \left( \therefore \frac { 1 }{ 125 } <1 \right) =5\left[ 1+\frac { 1 }{ 3 } (0.008) \right] =5(1+0.002666)=5.01\)

The general term in the expansion of (1 + x)^m is T_r+1 = ⁿC_r(1)^m-r x^r
On putting r = 2, we get T₃ = ^mC₂(1)^m-2 x² =  ^mC₂ x²
∴ Coefficient of x² = ^mC₂
Also, coefficient of x² in the expansion of (1+x)^m is 6
∴ ^mC₂ = 6 ⇒ \(\frac{m(m-1)}{2.1}=6 \Rightarrow m(m-1)=12\)
⇒ m(m-1) = 4.3
⇒ m = 4 

Since H is the H. M. between a and b,
\(we\quad get\quad H=\frac { 2ab }{ a+b } \quad ........(1)\)
\(LHS=(H-2a)(H-2b)\)
\(=\left( \frac { 2ab }{ a+b } -2a \right) \left( \frac { 2ab }{ a+b } -2b \right) \)
\(=\left( \frac { 2ab-2{ a }^{ 2 }-2ab }{ a+b } \right) \left( \frac { 2ab--2ab-{ ab }^{ 2 } }{ a+b } \right) \)
\(=\left( \frac { { -2a }^{ 2 } }{ a+b } \right) \left( \frac { { -2b }^{ 2 } }{ a+b } \right) =\left( \frac { { 4a }^{ 2 }{ b }^{ 2 } }{ { \left( a+b \right) }^{ 2 } } \right) { \left( \frac { 2ab }{ a+b } \right) }^{ 2 }\)
\(={ H }^{ 2 }[using\quad 1]\)

Given a, b, c are in A.P
\(\Rightarrow b=\frac { a+c }{ 2 } \)
RHS = 4[b² - ac]
\(=4\left[ { \left( \frac { a+c }{ 2 } \right) }^{ 2 }-ac \right] =4\left[ { \left( \frac { a+c }{ 4 } \right) }^{ 2 }-ac \right] \)
\(=4\left[ \frac { { \left( a+c \right) }^{ 2 }-4ac }{ 4 } \right] ={ a }^{ 2 }+{ c }^{ 2 }+2ac-4ac\)
= a² + c² - 2ac
= (a - c)² = LHS
Hence proved.

\((1+x)^{ m }=1+mx+\frac { m(m-1) }{ 2! } { x }^{ 2 }+..\)    [Binomial theorem for rational index]
∴ Co-efficient of x² = \(\frac { m(m-1) }{ 2 } \)
Given \(\frac { m(m-1) }{ 2 } \) = 6 ⇒ m²- m = 12
⇒ m²-m -12 = 0 ⇒ (m - 4) (m + 3) = 0
⇒ m = 4 or -3
∴ Negative value of m is -3