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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 ✓
a+1=x,b+1=y,c+1=zthenxyz=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=−6then(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✓
Commented by mr W last updated on 06/Jul/22
19^3 −111=6748 ✓
193−111=6748✓
Commented by Shrinava last updated on 06/Jul/22
Yes professor, thank you so much
Yesprofessor,thankyousomuch
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
{(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=3−14(a+b+c)+2(ab+bc+ca)+abc=−2−89(a+b+c)+3(ab+bc+ca)+abc=−1−27a+b+c=x,ab+bc+ca=y,abc=z{x+y+z=24x+2y+z=−109x+3y+z=−28x=−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
Commented by mathlove last updated on 06/Jul/22
thanks sir
thankssir

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