Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

$${thanks}\:{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com