Math, asked by moon1237879, 1 year ago

Please solve this problem! (a^2+b^2):ab=25:12 find the value of (a^3+b^3):(a^3-b^3)

Answers

Answered by Anant02
15
i hope it will help you
Attachments:

moon1237879: but the answer is only 91/37
moon1237879: never take proportions directly
moon1237879: always take constant
moon1237879: a:b=4:3 then a=4k and b=3k
moon1237879: the last portion is not correct
Anant02: here i use componendo and dividendo rule
moon1237879: the answer is correct
Answered by vinod04jangid
8

Answer:

The value of (a^3+b^3):(a^3-b^3)=  \frac{91}{37} \text { or } \frac{91}{-37}

Step-by-step explanation:

Given: (a^2+b^2):ab=25:12

To find the value of (a^3+b^3):(a^3-b^3).

\frac{a^{2}+b^{2}}{a b}=\frac{25}{12}

\frac{a^{2}+b^{2}}{2 a b}=\frac{25}{24}

Using componendo & dividendo we have:

\frac{a^{2}+b^{2}+2 a b}{a^{2}+b^{2}-2 a b}=\frac{25+24}{25-24}

\frac{(a+b)^{2}}{(a-b)^{2}}=\frac{49}{1}

\frac{a+b}{a-b}= \frac{\pm7}{1}

Again using componendo & dividendo we have:

\frac{a+b+a-b}{a+b-a+b}=\frac{\pm7+1}{\pm7-1}

&\frac{2 a}{2 b}=\frac{8}{6} \text { or } \frac{-6}{-8}

\frac{a}{b}=\frac{4}{3} \text { or } \frac{3}{4}

\frac{a^{3}}{b^{3}}=\frac{64}{27} \text { or } \frac{27}{64}

Again using componendo & dividendo

\frac{a^{3}+b^{3}}{a^{3}-b^{3}}=\frac{64+27}{64-27} \text { or } \frac{27+64}{27-64}

\frac{91}{37} \text { or } \frac{91}{-37}

#SPJ3

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