Question

write the square binomial (3x-1/3x)²​

Answer 1

Answer:

$\red {\left( 3x - \frac{1}{3x}\right)^{2}}$

$\green {= 9x^{2} - 2 + \frac{1}{9x^{2}}}$

Step-by-step explanation:

$\left( 3x - \frac{1}{3x}\right)^{2}$

$= (3x)^{2} - 2\times 3x \times \left(\frac{1}{3x}\right) + \left(\frac{1}{3x}\right)^{2}$

$\boxed { \pink { (a-b)^{2} = a^{2}-2ab+b^{2}}}$

$= 9x^{2} - 2 + \frac{1}{9x^{2}}$

Therefore.,

$\red {\left( 3x - \frac{1}{3x}\right)^{2}}$

$\green {= 9x^{2} - 2 + \frac{1}{9x^{2}}}$

•••♪

Answer 2

Step-by-step explanation:

(3x−

3x

1

)

2

\green {= 9x^{2} - 2 + \frac{1}{9x^{2}}}=9x

2

−2+

9x

2

1

Step-by-step explanation:

\left( 3x - \frac{1}{3x}\right)^{2}(3x−

3x

1

)

2

= (3x)^{2} - 2\times 3x \times \left(\frac{1}{3x}\right) + \left(\frac{1}{3x}\right)^{2}=(3x)

2

−2×3x×(

3x

1

)+(

3x

1

)

2

\boxed { \pink { (a-b)^{2} = a^{2}-2ab+b^{2}}}

(a−b)

2

=a

2

−2ab+b

2

= 9x^{2} - 2 + \frac{1}{9x^{2}}=9x

2

−2+

9x

2

1

Therefore.,

\red {\left( 3x - \frac{1}{3x}\right)^{2}}(3x−

3x

1

)

2

\green {= 9x^{2} - 2 + \frac{1}{9x^{2}}}=9x

2

−2+

9x

2

1

•••♪