Question

(1/√3)² + √3² + 2 = (2/√3)² x 4

PROVE LHS=RHS

$\frac{1}{\sqrt{3}}^{2} + \sqrt{3}^{2} +2 = \frac{2}{\sqrt{3}}^{2}$ × $4$

Answer 1

Step-by-step explanation:

$\frac{1}{9} + 9 + 2 = \frac{4}{9} \times 4$

$\frac{1 + 81 + 18}{9} = \frac{16}{9}$

$\frac{100}{9} = \frac{16}{9}$

Hence, this not proven LHS= RHS

Answer 2

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm :\longmapsto\: {\bigg[\dfrac{1}{ \sqrt{3} } \bigg]}^{2} + {( \sqrt{3} )}^{2} + 2$

$\rm \: = \: \dfrac{1}{3} + 3 + 2$

$\rm \: = \: \dfrac{1}{3} + 5$

$\rm \: = \: \dfrac{1}{3} + \dfrac{5}{1}$

$\rm \: = \: \dfrac{1 + 5 \times 3}{3}$

$\rm \: = \: \dfrac{1 + 15}{3}$

$\rm \: = \: \dfrac{16}{3}$

Hence,

$\rm :\longmapsto\: \boxed{ \tt{ \: {\bigg[\dfrac{1}{ \sqrt{3} } \bigg]}^{2} + {( \sqrt{3} )}^{2} + 2 = \frac{16}{3} \: }} - - - - (1)$

Now, Consider RHS

$\rm :\longmapsto\: {\bigg[\dfrac{2}{ \sqrt{3} } \bigg]}^{2} \times 4$

$\rm \: = \: \dfrac{4}{3} \times 4$

$\rm \: = \: \dfrac{16}{3}$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: {\bigg[\dfrac{2}{ \sqrt{3} } \bigg]}^{2} \times 4 = \frac{16}{3} \: }} - - - (2)$

From equation (1) and equation (2), we concluded that

$\rm :\longmapsto\: \boxed{ \tt{ \: {\bigg[\dfrac{1}{ \sqrt{3} } \bigg]}^{2} + {( \sqrt{3} )}^{2} + 2 = {\bigg[\dfrac{2}{ \sqrt{3} } \bigg]}^{2} \times 4 \: }}$

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More Identities to know :-

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)