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Question Number 181593 by mnjuly1970 last updated on 27/Nov/22

I f , a^( 2) +5 b^( 2) + 4c^( 2) = 4b (a +c ) then , find the value of : E = (( ( b+ c −a )^( 3) )/( abc)) = ? ( abc ≠ 0 )

$$ \\ $$$$\:\:\:\:\:\:\:\mathrm{I}\:\mathrm{f}\:,\:\:\:\:{a}^{\:\mathrm{2}} \:+\mathrm{5}\:{b}^{\:\mathrm{2}} \:+\:\mathrm{4}{c}^{\:\mathrm{2}} =\:\mathrm{4}{b}\:\left({a}\:+{c}\:\right) \\ $$$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\mathrm{then}\:,\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:: \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{E}\:=\:\frac{\:\left(\:{b}+\:{c}\:−{a}\:\right)^{\:\mathrm{3}} }{\:{abc}}\:=\:?\:\:\:\:\left(\:{abc}\:\neq\:\mathrm{0}\:\right)\:\:\:\:\:\:\: \\ $$

Answered by mahdipoor last updated on 27/Nov/22

⇒a^2 +5b^2 +4c^2 −4ba−4bc=0 ⇒(a^2 +4b^2 −4ba)+(b^2 +4c^2 −4bc)=0⇒ (a−2b)^2 +(b−2c)^2 =0⇒ { ((a−2b=0⇒a=2b)),((b−2c=0⇒b=2c)) :} E=(((b+c−a)^3 )/(abc))=(((2c+c−4c)^3 )/((4c)(2c)(c)))=((−c^3 )/(8c^3 ))=(1/8)

$$\Rightarrow{a}^{\mathrm{2}} +\mathrm{5}{b}^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} −\mathrm{4}{ba}−\mathrm{4}{bc}=\mathrm{0} \\ $$$$\Rightarrow\left({a}^{\mathrm{2}} +\mathrm{4}{b}^{\mathrm{2}} −\mathrm{4}{ba}\right)+\left({b}^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} −\mathrm{4}{bc}\right)=\mathrm{0}\Rightarrow \\ $$$$\left({a}−\mathrm{2}{b}\right)^{\mathrm{2}} +\left({b}−\mathrm{2}{c}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow\begin{cases}{{a}−\mathrm{2}{b}=\mathrm{0}\Rightarrow{a}=\mathrm{2}{b}}\\{{b}−\mathrm{2}{c}=\mathrm{0}\Rightarrow{b}=\mathrm{2}{c}}\end{cases} \\ $$$${E}=\frac{\left({b}+{c}−{a}\right)^{\mathrm{3}} }{{abc}}=\frac{\left(\mathrm{2}{c}+{c}−\mathrm{4}{c}\right)^{\mathrm{3}} }{\left(\mathrm{4}{c}\right)\left(\mathrm{2}{c}\right)\left({c}\right)}=\frac{−{c}^{\mathrm{3}} }{\mathrm{8}{c}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{8}} \\ $$

Commented by mnjuly1970 last updated on 27/Nov/22

mamnoon jenab mahdipoor zendeh bashid.

$$\:\:{mamnoon}\:{jenab}\:{mahdipoor} \\ $$$${zendeh}\:{bashid}. \\ $$

Answered by madhabnandi last updated on 27/Nov/22

Ans: a^2 +5b^2 +4c^2 =4b(a+c) ⇒ a^2 +4b^2 +b^2 +4c^2 =4ab+4bc ⇒ a^2 −4ab+4b^2 + b^2 −4bc+4c^2 =0 ⇒ (a −2b)^2 + (b−2c)^2 =0 ∴ a−2b=0 and b−2c=0 ∴ a=2b ⇒b=2c . =4c . E = (( ( b+ c −a )^( 3) )/( abc)) =(((2c+c−4c)^3 )/(4c×2c×c)) =(((−c)^3 )/(8c^3 )) =((−c^3 )/(8c^3 )) =−(1/8) .

$$\:\:\:{Ans}:\: \\ $$$$\:\:\:\:\:\:{a}^{\mathrm{2}} +\mathrm{5}{b}^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} \:=\mathrm{4}{b}\left({a}+{c}\right) \\ $$$$\:\:\Rightarrow\:{a}^{\mathrm{2}} \:+\mathrm{4}{b}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} =\mathrm{4}{ab}+\mathrm{4}{bc} \\ $$$$\:\:\Rightarrow\:{a}^{\mathrm{2}} \:−\mathrm{4}{ab}+\mathrm{4}{b}^{\mathrm{2}} \:+\:\:{b}^{\mathrm{2}} −\mathrm{4}{bc}+\mathrm{4}{c}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\Rightarrow\:\left({a}\:−\mathrm{2}{b}\right)^{\mathrm{2}} \:+\:\left({b}−\mathrm{2}{c}\right)^{\mathrm{2}} \:=\mathrm{0} \\ $$$$\therefore\:{a}−\mathrm{2}{b}=\mathrm{0}\:\:\:{and}\:\:\:{b}−\mathrm{2}{c}=\mathrm{0} \\ $$$$\therefore\:{a}=\mathrm{2}{b}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{b}=\mathrm{2}{c}\:. \\ $$$$\:\:\:\:\:\:\:=\mathrm{4}{c}\:. \\ $$$$\:\:\:\:\mathrm{E}\:=\:\frac{\:\left(\:{b}+\:{c}\:−{a}\:\right)^{\:\mathrm{3}} }{\:{abc}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{2}{c}+{c}−\mathrm{4}{c}\right)^{\mathrm{3}} }{\mathrm{4}{c}×\mathrm{2}{c}×{c}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\left(−{c}\right)^{\mathrm{3}} }{\mathrm{8}{c}^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{−{c}^{\mathrm{3}} }{\mathrm{8}{c}^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{8}}\:. \\ $$$$\:\:\: \\ $$

Commented by mnjuly1970 last updated on 27/Nov/22

grateful sir

$${grateful}\:{sir} \\ $$

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