Question

$\fcolorbox{red}{orange}{PLEASE SOLVE THIS QUESTION}$
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Answer 1

Explanation:

1. 15a² + 4b² + ( 15a² - 4b² ) / 15a² + 4b² - ( 15a² - 4b² ) = 47 + 7 / 47 - 7 [ By componendo dividends]

15a² + 4b² + 15a² - 4b² / 15a² + 4b² - 15a² + 4b² = 54 / 40

30a² / 8b² = 54 / 40

a²/ b² = 54 × 8 / 40 × 30

a² / b² = 9 / 25

a² / b² = 3/5 [ Take positive square root of both sides ]

a² / b² = 3:5

To find ,

b³ - 2a³ / b³ + 2a³

a/ b = 3/5

[ a/b ] ³ = [ 3/5 ] ³

a/ b = 27 / 125 [ by cubing both ]

so ,

( a/b)³ = 27 / 125

b³ / a³ = 125 / 27 [ BY INVERTENDO ]

b³/ a³ × 1/2 = 125 / 27 × 1/2 ( multiplying both sides by 1/2 )

b² / 2a³ = 125 / 54

b³ + 2a³ / b³ -2a³ = 125 + 54 / 125 - 54

b³ + 2a³ / b³ - 2a³ = 179 / 71

b³ - 2a³ / b³ + 2a³ = 71 / 179 [ by invertendo ]

b³ - 2a² / b³ + 2a³ = 71 : 179

hõpê ít hélps yôú

Answer 2

Explanation:

Given that

• 15a²+4b²/15a² - 4b² = 47/7

To Find

• find value of the ratio b³-2a³/b³+2a³

Solution

$\tt \: \dfrac{15 {}^{2} + 4b {}^{2} }{ {15a}^{2} - {4b}^{2} } = \dfrac{47}{7} \qquad \qquad \: \green{given}$

$\longrightarrow \sf \: \frac{ {15a}^{2} + {4b}^{2} + ( {15a}^{2} {4b}^{2} )}{ {15a}^{2} - {4b}^{2} - ( {15a}^{2} - {4b}^{2} )} = \frac{47 + 7}{47 - 7} \\ \\ \\ \qquad \qquad \qquad \qquad \ \red{ \{{ \bf \: by \: componedo - dividendo \}}}$

$\longrightarrow \sf \: \: \frac{ {15a + {4b}^{2} } + {15b}^{2} - {4b}^{2} }{ {15b}^{2} + {4b}^{2} - {15b}^{2} {4b}^{2} } = \frac{54}{40} \\ \\ \\ \longrightarrow \sf \frac{ {15b {}^{2} + {15b}^{2} }^{} + {4b}^{2} - {4b}^{2} }{ {15b}^{2} - {15b}^{2} + {4b}^{2} + {4b}^{2} } = \frac{54}{40} \\ \\ \\ \longrightarrow \sf \frac{ {30a}^{2} }{ {8b}^{2} } = \frac{54}{40} \\ \\ \\ \longrightarrow \sf \frac{ {a}^{2} }{ {b}^{2} } = \frac{54 \times 8}{40 \times 30} \\ \\ \\ \longrightarrow \sf \frac{ {a}^{2} }{ {b}^{2} } = \frac{9}{25} \\ \\ \\ \longrightarrow \sf \frac{a}{b} = \frac{ \sqrt{9} }{ \sqrt{25} } \\ \\ \\ \longrightarrow \sf \frac{a}{b} = \frac{ \sqrt{3 \times 3} }{ \sqrt{5 \times 5} }$

$\qquad \longrightarrow \sf \: \frac{a}{b} = \frac{3}{5} \\ \\ \\ \qquad \qquad \qquad \qquad \red{ \{ \bf \: taking \: positive \: square \: root \: of \: both \: sides \}}$

now,

$\bf \: we \: find \: the \: value \: of \: the \: \frac{ {b}^{3} - {2a}^{3} }{ {b}^{3} + {2a}^{3} }$

$\qquad \qquad \qquad \sf \frac{a}{b} = \frac{3}{5}$

$\longrightarrow \sf \: \: \bigg \lgroup \frac{a}{b} \bigg \rgroup ^{3} = \bigg \lgroup\frac{3}{5} \bigg \rgroup ^{3} \\ \qquad \qquad \red{\{ \bf \: cubing \: both \: sides \}}$

$\sf \: so$

$\longrightarrow \sf \frac{ {a}^{3} }{ {b}^{3} } = \frac{27}{125} \\ \\ \\ \longrightarrow \sf \frac{ {b}^{3} }{ {a}^{3} } = \frac{125}{27} \qquad \qquad \red{\{ \bf \: by \: invertendo \}}$

$\longrightarrow \sf \: \frac{ {b}^{3} }{ {a}^{ 3} } \times \frac{1}{3} = \frac{125}{27} \times \frac{1}{3} \\ \\ \\ \red{\bf \{ multipying \: both \: side \: by \: \frac{1}{3} \}}$

$\longrightarrow \sf \frac{ {b}^{3} }{ {a}^{3} } = \frac{125}{54} \\ \\ \\ \longrightarrow \sf \frac{ {b}^{3} }{ {2a}^{3} } = \frac{125}{54}$

$\longrightarrow \sf \frac{ {b}^{3} + {2a}^{3} }{ {a}^{3} - {2a}^{3} } = \frac{125 + 54}{125 - 54} \\ \\ \red{ \{ \bf \: by \: componendo - \: dividend \}}$

$\longrightarrow \sf \frac{ {b}^{3} + {2a}^{3} }{ {b}^{3} - {2a}^{3} } = \frac{179}{71}$

$\longrightarrow \sf \frac{ {b}^{3} - {2a}^{3} }{ {b}^{3} + {2a}^{3} } = \frac{71}{179} \qquad \qquad \red {\{ \bf \: by \: invertendo \}}$

$\qquad \sf \therefore \qquad \boxed{ \frak {\frac{ {b}^{3} - {2a}^{3} }{ {b}^{3} + {2a}^{3} } = 71 : 179 }}$

$\qquad \sf \:the \: ratio \: value \qquad { {\frac{ {b}^{3} - {2a}^{3} }{ {b}^{3} + {2a}^{3} } \: \: \: be \: \: \: 71 : 179 }}$