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Question-173054

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Question Number 173054 by mathlove last updated on 06/Jul/22
Commented by Shrinava last updated on 06/Jul/22
a+1=x , b+1=y , c+1=z then xyz=3 ⇒ x+y+z+xy+yz+zx=−6 (i) ⇒ (x+2)(y+2)(z+2)=−1 ⇒ 4x+4y+4z+2xy+2yz+2zx=−12 ⇒ 2(x+y+z)+xy+yz+zx=−6 (ii) (i) and (ii) ⇒ x+y+z = 2(x+y+z) ⇒ x+y+z=0 , xyz=3 , xy+yz+zx=−6 then (a+20)(b+20)(c+20)=(x+20)(y+20)(z+20) ⇒ 36(x+y+z)+19(xy+yz+zx)+xyz ⇒ 0 + 19 ∙ (−6) + 3 = −111 ✓
$$\mathrm{a}+\mathrm{1}=\mathrm{x}\:\:,\:\:\mathrm{b}+\mathrm{1}=\mathrm{y}\:\:,\:\:\mathrm{c}+\mathrm{1}=\mathrm{z} \\ $$$$\mathrm{then}\:\:\:\mathrm{xyz}=\mathrm{3} \\ $$$$\Rightarrow\:\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{xy}+\mathrm{yz}+\mathrm{zx}=−\mathrm{6}\:\:\:\left(\mathrm{i}\right) \\ $$$$\Rightarrow\:\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{y}+\mathrm{2}\right)\left(\mathrm{z}+\mathrm{2}\right)=−\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{4x}+\mathrm{4y}+\mathrm{4z}+\mathrm{2xy}+\mathrm{2yz}+\mathrm{2zx}=−\mathrm{12} \\ $$$$\Rightarrow\:\mathrm{2}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)+\mathrm{xy}+\mathrm{yz}+\mathrm{zx}=−\mathrm{6}\:\:\:\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{i}\right)\:\:\mathrm{and}\:\:\left(\mathrm{ii}\right)\:\:\Rightarrow \\ $$$$\mathrm{x}+\mathrm{y}+\mathrm{z}\:=\:\mathrm{2}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right) \\ $$$$\Rightarrow\:\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{0}\:\:,\:\:\mathrm{xyz}=\mathrm{3}\:\:,\:\:\mathrm{xy}+\mathrm{yz}+\mathrm{zx}=−\mathrm{6} \\ $$$$\mathrm{then}\:\:\left(\mathrm{a}+\mathrm{20}\right)\left(\mathrm{b}+\mathrm{20}\right)\left(\mathrm{c}+\mathrm{20}\right)=\left(\mathrm{x}+\mathrm{20}\right)\left(\mathrm{y}+\mathrm{20}\right)\left(\mathrm{z}+\mathrm{20}\right) \\ $$$$\Rightarrow\:\mathrm{36}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)+\mathrm{19}\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)+\mathrm{xyz} \\ $$$$\Rightarrow\:\mathrm{0}\:+\:\mathrm{19}\:\centerdot\:\left(−\mathrm{6}\right)\:+\:\mathrm{3}\:=\:−\mathrm{111}\:\:\checkmark \\ $$
Commented by mr W last updated on 06/Jul/22
19^3 −111=6748 ✓
$$\mathrm{19}^{\mathrm{3}} −\mathrm{111}=\mathrm{6748}\:\checkmark \\ $$
Commented by Shrinava last updated on 06/Jul/22
Yes professor, thank you so much
$$\mathrm{Yes}\:\mathrm{professor},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$
Answered by Rasheed.Sindhi last updated on 06/Jul/22
{ (((a+1)(b+1)(c+1)=3)),(((a+2)(b+2)(c+2)=−2)),(((a+3)(b+3)(c+3)=−1)),(((a+20)(b+20)(c+20)=?)) :} { (((a+b+c)+(ab+bc+ca)+abc=3−1)),((4(a+b+c)+2(ab+bc+ca)+abc=−2−8)),((9(a+b+c)+3(ab+bc+ca)+abc=−1−27)) :} a+b+c=x, ab+bc+ca=y, abc=z { ((x+y+z=2)),((4x+2y+z=−10)),((9x+3y+z=−28)) :} x=−3,y=−3,z=8 (a+20)(b+20)(c+20) =8000+400(a+b+c)+20(ab+bc+ca)+abc =8000+400x+20y+z =8000+400(−3)+20(−3)+(8) =8000−1200−60+8 =6748
$$\begin{cases}{\left({a}+\mathrm{1}\right)\left({b}+\mathrm{1}\right)\left({c}+\mathrm{1}\right)=\mathrm{3}}\\{\left({a}+\mathrm{2}\right)\left({b}+\mathrm{2}\right)\left({c}+\mathrm{2}\right)=−\mathrm{2}}\\{\left({a}+\mathrm{3}\right)\left({b}+\mathrm{3}\right)\left({c}+\mathrm{3}\right)=−\mathrm{1}}\\{\left({a}+\mathrm{20}\right)\left({b}+\mathrm{20}\right)\left({c}+\mathrm{20}\right)=?}\end{cases}\: \\ $$$$\begin{cases}{\left({a}+{b}+{c}\right)+\left({ab}+{bc}+{ca}\right)+{abc}=\mathrm{3}−\mathrm{1}}\\{\mathrm{4}\left({a}+{b}+{c}\right)+\mathrm{2}\left({ab}+{bc}+{ca}\right)+{abc}=−\mathrm{2}−\mathrm{8}}\\{\mathrm{9}\left({a}+{b}+{c}\right)+\mathrm{3}\left({ab}+{bc}+{ca}\right)+{abc}=−\mathrm{1}−\mathrm{27}}\end{cases} \\ $$$${a}+{b}+{c}={x},\:{ab}+{bc}+{ca}={y},\:{abc}={z} \\ $$$$\begin{cases}{{x}+{y}+{z}=\mathrm{2}}\\{\mathrm{4}{x}+\mathrm{2}{y}+{z}=−\mathrm{10}}\\{\mathrm{9}{x}+\mathrm{3}{y}+{z}=−\mathrm{28}}\end{cases} \\ $$$$\:\:{x}=−\mathrm{3},{y}=−\mathrm{3},{z}=\mathrm{8} \\ $$$$ \\ $$$$\left({a}+\mathrm{20}\right)\left({b}+\mathrm{20}\right)\left({c}+\mathrm{20}\right) \\ $$$$\:\:\:=\mathrm{8000}+\mathrm{400}\left({a}+{b}+{c}\right)+\mathrm{20}\left({ab}+{bc}+{ca}\right)+{abc} \\ $$$$\:\:=\mathrm{8000}+\mathrm{400}{x}+\mathrm{20}{y}+{z} \\ $$$$\:\:=\mathrm{8000}+\mathrm{400}\left(−\mathrm{3}\right)+\mathrm{20}\left(−\mathrm{3}\right)+\left(\mathrm{8}\right) \\ $$$$\:\:\:=\mathrm{8000}−\mathrm{1200}−\mathrm{60}+\mathrm{8} \\ $$$$\:\:\:=\mathrm{6748} \\ $$
Commented by mathlove last updated on 06/Jul/22
thanks sir
$${thanks}\:{sir} \\ $$

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