343 =73
So (343)^2/3 = (7³)^2/3 = \({ 7 }^{ 7\times \frac { 2 }{ 3 } }\) = 7² = 49

\(\frac{1}{6(x-1)}-\frac{1}{2(x+1)}+\frac{1}{3(x+2)}\)

1.8572

\(\sqrt{5}-\sqrt{4}\)

\(\sqrt{5}-\sqrt{4}\)

x² - 2x - 5 = 0

x = 1 or \(\frac{3}{2}\)

Given \(\frac { 2-\sqrt { 3 } }{ \sqrt { 3 } } \)
The rationalising factor of the denominator is \(\sqrt { 3 } \)
\(\therefore \frac { 2-\sqrt { 3 } }{ \sqrt { 3 } } =\frac { 2-\sqrt { 3 } }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } =\frac { 2\sqrt { 3 } -3 }{ \sqrt { 3 } } \)
= \(\frac { 2(1.732)-3 }{ 3 } =\frac { 3.464-3 }{ 3 } =\frac { 0.464 }{ 3 } \) = 0.154

Given a³+b³ = ab (8-3a-3b)
⇒ a³+ b³ = 8ab - 3a²b - 3ab²
⇒  a³+b³+3a²b+3ab²=8ab
⇒ (a+b)³= 2³ ab
⇒ \(\left( \frac { a+b }{ 2 } \right) ^{ 3 }\)= ab
⇒  log\(\left( \frac { a+b }{ 2 } \right) \) = log ab    [Taking logarithm on both sides]
⇒ 3 log\(\left( \frac { a+b }{ 2 } \right) \)  log a + log b   [Using power rule and products rule]
⇒  log \(\left( \frac { a+b }{ 2 } \right) \) = \(\frac { 1 }{ 3 } \) [log a + log b]

Here a = 9,  b = 5, c = 0
\(\therefore\)  D = b² - 4ac = 52 - 4 (9) (0) = 25
D > 0 and it is a perfect square, the roots are real and distinct

Here a = 4, b = -1, c = -2
\(\therefore\) D = b²- 4ac = (-1)² - 4 (4) (-2)
= 1 + 32 = 33
Since D > 0, the two roots are real and distinct.

Given |x| - 10 < -3
⇒ |x| < -3 + 1
⇒ |x| < 7
This means -7 < x < 7.​​​​​​
\(\therefore\) The Solution set is (-7, 7)

LHS = \(log\frac { { a }^{ 2 } }{ bc } +log\frac { b^{ 2 } }{ ca } +log\frac { c^{ 2 } }{ ab } \)
= \(log\left( \frac { { a }^{ 2 } }{ bc } .\frac { b^{ 2 } }{ ca } .\frac { c^{ 2 } }{ ab } \right) \)
= \(log\left( \frac { { a }^{ 2 }b^{ 2 }c^{ 2 } }{ { a }^{ 2 }b^{ 2 }c^{ 2 } } \right) \)
= log 1 = 0 =RHS
Hence proved

Given \({ log }_{ 9 }^{ 27 }-{ log }_{ 27 }^{ 9 }\)
= \({ log }_{ 9 }^{ 3^3 }-{ log }_{ 27 }^{ 3^2 }\)
=  3 log₉3-2 log₂₇3  [By power rule]
=   \(\frac { 3 }{ { log }_{ 3 }9 } -\frac { 2 }{ { log }_{ 3 }27 } \)    [By change of the base rule]
= \(\frac { 3 }{ { log }_{ 3 }{ 3 }^{ 2 } } -\frac { 2 }{ { log }_{ 3 }{ 3 }^{ 3 } } \)
= \(\frac { 3 }{ { 2\quad log }_{ 3 }{ 3 } } -\frac { 2 }{ { 3\quad log }_{ 3 }{ 3 } } \)
=  \(\frac { 3 }{ 2 } -\frac { 2 }{ 3 } \)      [∵ log₃³ = 1]
= \(\frac { 9-4 }{ 6 } \)

Given -3|x| + 5 < -2
⇒ -3 Ixl < -2 - 5
⇒ -3|x| < -7
\(⇒\ |x|\ge {7\over 3}\)
This means \(-{7\over 3}\ge\ |x|\ge\ {7\over 3}\)

[Dividing by -3]
∴ The solution set \(\left( -\infty,\cup{-7\over 3}]\cup[{7\over 3},\infty \right)\)