Menu Close

Question-187989

Tinku Tara 0 Comments
Pin




Question Number 187989 by mnjuly1970 last updated on 24/Feb/23
Answered by HeferH last updated on 24/Feb/23
b^2 =ac b^3 =abc ((b^3 c^3 +a^3 c^3 +a^3 b^3 )/(a^3 b^3 c^3 ))∙(((a^2 b^2 c^2 )/(a^3 +b^3 +c^3 )))= ((b^3 c^3 +a^3 c^3 +a^3 b^3 )/(abc))∙((1/(a^3 +b^3 +c^3 )))= (((bc)^3 +b^6 +(ab)^3 )/b^3 )∙((1/(a^3 +b^3 +c^3 )))= c^3 + b^3 + a^3 ∙((1/(a^3 +b^3 +c^3 )))= 1
$$\:{b}^{\mathrm{2}} ={ac} \\ $$$$\:{b}^{\mathrm{3}} ={abc} \\ $$$$\:\frac{{b}^{\mathrm{3}} {c}^{\mathrm{3}} +{a}^{\mathrm{3}} {c}^{\mathrm{3}} +{a}^{\mathrm{3}} {b}^{\mathrm{3}} }{{a}^{\mathrm{3}} {b}^{\mathrm{3}} {c}^{\mathrm{3}} }\centerdot\left(\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }\right)= \\ $$$$\:\frac{{b}^{\mathrm{3}} {c}^{\mathrm{3}} +{a}^{\mathrm{3}} {c}^{\mathrm{3}} +{a}^{\mathrm{3}} {b}^{\mathrm{3}} }{{abc}}\centerdot\left(\frac{\mathrm{1}}{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }\right)= \\ $$$$\:\frac{\left({bc}\right)^{\mathrm{3}} +{b}^{\mathrm{6}} +\left({ab}\right)^{\mathrm{3}} }{{b}^{\mathrm{3}} }\centerdot\left(\frac{\mathrm{1}}{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }\right)= \\ $$$$\:{c}^{\mathrm{3}} +\:{b}^{\mathrm{3}} \:+\:{a}^{\mathrm{3}} \centerdot\left(\frac{\mathrm{1}}{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }\right)=\:\mathrm{1} \\ $$$$\: \\ $$
Answered by MathGuy last updated on 24/Feb/23
Answer :− ((a^2 b^2 c^2 )/(a^3 +b^3 +c^3 ))(((b^3 c^3 +c^3 a^3 +a^3 b^3 )/(a^3 b^3 c^3 ))) ⇒ (1/(a^3 +b^3 +c^3 ))(((b^3 c^3 +c^3 a^3 +a^3 b^3 )/(abc))) ⇒ (1/(a^3 +b^3 +c^3 ))(((b^2 c^2 )/a)+((c^2 a^2 )/b)+((a^2 b^2 )/c)) substitute b^2 =ac & c^2 a^2 =b^2 , in the expression under brackets and simplify. you will get (1/(a^3 +b^3 +c^3 ))(a^3 +b^3 +c^3 ) = 1
$${Answer}\::−\: \\ $$$$\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }\left(\frac{{b}^{\mathrm{3}} {c}^{\mathrm{3}} +{c}^{\mathrm{3}} {a}^{\mathrm{3}} +{a}^{\mathrm{3}} {b}^{\mathrm{3}} }{{a}^{\mathrm{3}} {b}^{\mathrm{3}} {c}^{\mathrm{3}} }\right) \\ $$$$ \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }\left(\frac{{b}^{\mathrm{3}} {c}^{\mathrm{3}} +{c}^{\mathrm{3}} {a}^{\mathrm{3}} +{a}^{\mathrm{3}} {b}^{\mathrm{3}} }{{abc}}\right) \\ $$$$ \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }\left(\frac{{b}^{\mathrm{2}} {c}^{\mathrm{2}} }{{a}}+\frac{{c}^{\mathrm{2}} {a}^{\mathrm{2}} }{{b}}+\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{c}}\right) \\ $$$${substitute}\:{b}^{\mathrm{2}} ={ac}\:\&\:{c}^{\mathrm{2}} {a}^{\mathrm{2}} ={b}^{\mathrm{2}} ,\:{in}\:{the}\:{expression} \\ $$$${under}\:{brackets}\:{and}\:{simplify}. \\ $$$${you}\:{will}\:{get}\: \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right) \\ $$$$=\:\mathrm{1}\: \\ $$$$ \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *