Question

If a,b,c,d are in continued properties, prove that \frac{(a-b)^{3}}{(b-c)^{3}

Transformation properties
ANSWERED
asked 2021-08-18
If \(\displaystyle{a},{b},{c},{d}\) are in continued properties, prove that
\(\displaystyle{\frac{{{\left({a}-{b}\right)}^{{{3}}}}}{{{\left({b}-{c}\right)}^{{{3}}}}}}={\frac{{{a}}}{{{d}}}}\)

Expert Answers (1)

2021-08-19
Step 1
\(\displaystyle{a},{b},{c},{d}\) are in continued proportion
Therefore, we have
\(\displaystyle{\frac{{{a}}}{{{b}}}}={\frac{{{b}}}{{{c}}}}={\frac{{{c}}}{{{d}}}}={k}{\left({s}{a}{y}\right)}\)
\(\displaystyle{c}={d}{k},\ {b}={c}{k}={d}{k}^{{{2}}},\ {a}={b}{k}={d}{k}^{{{3}}}\)
Step 2
Consider,
\(\displaystyle{\frac{{{\left({a}-{b}\right)}^{{{3}}}}}{{{\left({b}-{c}\right)}^{{{3}}}}}}={\frac{{{\left({d}{k}^{{{3}}}-{d}{k}^{{{2}}}\right)}^{{{3}}}}}{{{\left({d}{k}^{{{2}}}-{d}{k}\right)}^{{{3}}}}}}\)
\(\displaystyle={\frac{{{d}^{{{3}}}{k}^{{{6}}}{\left({k}-{1}\right)}}}{{{d}^{{{3}}}{k}^{{{3}}}{\left({k}-{1}\right)}}}}\)
\(\displaystyle={k}^{{{3}}}\)
\(\displaystyle{\frac{{{a}}}{{{d}}}}={\frac{{{d}{k}^{{{3}}}}}{{{d}}}}={k}^{{{3}}}\)
Hence, \(\displaystyle{\frac{{{\left({a}-{b}\right)}^{{{3}}}}}{{{\left({b}-{c}\right)}^{{{3}}}}}}={\frac{{{a}}}{{{d}}}}\)
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