Question

ab_bc+ac,bc_ca+ab,ca_ab_2bc​

Answer 1

Correct Question:-

${\dfrac {ab}{bc}}+ac={\dfrac {bc}{ca}}+ab={\dfrac {ca}{ab}}+bc$

Required Answer:-

${\dfrac {ab}{bc}}+ac={\dfrac {bc}{ca}}+ab={\dfrac {ca}{ab}}+bc$

${:}\dashrightarrow$$\sf {{\dfrac {ab+ac×bc}{ab}}={\dfrac {bc+ab×ca}{ca}}={\dfrac{ca+bc×ab}{ab}}}$

${:}\dashrightarrow$$\sf {{\dfrac {ab+abc {}^{2}}{bc}}={\dfrac {bc+a {}^{2}bc}{ca}}={\dfrac {ca+ab{}^{2}c}{ab}}}$

${:}\dashrightarrow$$\sf {{\dfrac {ab (1+c {}^{2})}{bc}}={\dfrac {bc (1+a {}^{2})}{ca}}={\dfrac {ca (1+b {}^{2})}{ab}}}$

• According to the formula

${\boxed{{\dfrac {a}{b}}={\dfrac {c}{d}}={\dfrac {e}{f}}\implies{ad=cf=be}}}$

${:}\dashrightarrow$$\sf {ca ×ab (1+c{}^{2})=ab×bc(1+a {}^{2})=bc×ca (1+b {}^{2})}$

${:}\dashrightarrow$$\sf {{a}^{2}bc(1+c {}^{2})=ab {}^{2}c (1+c {}^{2})=abc {}^{2}(1+b {}^{2})}$

${:}\dashrightarrow$$\sf {{\cancel {{a}^{2}bc}}+{\bcancel {a{}^{2}bc {}^{2}}}={\cancel{ab {}^{2}c}}+{\bcancel{abc {}^{2}}}={\cancel{abc {}^{2}}}+{\bcancel{ab {}^{2}c {}^{2}}}}$

${:}\dashrightarrow$$\sf {c={\cancel{bc}}={\cancel{bc}}}$

${:}\dashrightarrow$$\sf {c=bc}$

$\therefore$${\underline{\boxed{\bf {c=bc}}}}$