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Question Number 194522 by cortano12 last updated on 09/Jul/23

Answered by witcher3 last updated on 09/Jul/23

u=((x−5))^(1/3) v=((7−x))^(1/3) u^3 +v^3 =2p ((v−u)/(u+v))=(1/2)(v^3 −u^3 ) ⇒(v−u)((1/2)(v+u)(u^2 +v^2 +uv)−1)=0 v=u⇒x−5=7−x⇒x=6 (v+u)(u^2 +v^2 +uv)=2 u^3 +v^3 +2uv(u+v)=2 ⇔2+2uv(u+v)=0 ⇒uv(u+v)=0⇒u=0,v=0,u=−v u=0⇒x=5,v=0⇒x=7 u=−v⇒u^3 =−v^3 ⇒x−7=x−5..impossible S={5,7,6}

u=x−53v=7−x3u3+v3=2pv−uu+v=12(v3−u3)⇒(v−u)(12(v+u)(u2+v2+uv)−1)=0v=u⇒x−5=7−x⇒x=6(v+u)(u2+v2+uv)=2u3+v3+2uv(u+v)=2⇔2+2uv(u+v)=0⇒uv(u+v)=0⇒u=0,v=0,u=−vu=0⇒x=5,v=0⇒x=7u=−v⇒u3=−v3⇒x−7=x−5..impossibleS={5,7,6}

Answered by Rasheed.Sindhi last updated on 09/Jul/23

((((7−x))^(1/3) −((x−5))^(1/3) )/( ((7−x))^(1/3) +((x−5))^(1/3) ))=((6−x)/1) (a/b)=(c/d)⇔((a+b)/(a−b))=((c+d)/(c−d)) (((((7−x))^(1/3) −((x−5))^(1/3) )+(((7−x))^(1/3) +((x−5))^(1/3) ))/((((7−x))^(1/3) −((x−5))^(1/3) )−(((7−x))^(1/3) +((x−5))^(1/3) )))=(((6−x)+1)/((6−x)−1)) ((2((7−x))^(1/3) )/(−2((x−5))^(1/3) ))=((7−x)/(5−x)) ((((7−x))^(1/3) )/( ((x−5))^(1/3) ))=((7−x)/(x−5)) ((7−x )/( x−5))=(((7−x)/(x−5)))^3 (((7−x)/(x−5)))[(((7−x)/(x−5)))^2 −1]=0 ((7−x)/(x−5))=0 ∣ (((7−x)/(x−5)))^2 =1 x=7✓ ∣ 7−x=±(x−5) ∣ 7−x=x−5 ∣ 7−x=−x+5_(Absurd) ∣ x=6 ✓ I lost x=5 which is also a root!

7−x3−x−537−x3+x−53=6−x1ab=cd⇔a+ba−b=c+dc−d(7−x3−x−53)+(7−x3+x−53)(7−x3−x−53)−(7−x3+x−53)=(6−x)+1(6−x)−127−x3−2x−53=7−x5−x7−x3x−53=7−xx−57−xx−5=(7−xx−5)3(7−xx−5)[(7−xx−5)2−1]=07−xx−5=0∣(7−xx−5)2=1x=7✓∣7−x=±(x−5)∣7−x=x−5∣7−x=−x+5Absurd∣x=6✓Ilostx=5whichisalsoaroot!

Commented by BaliramKumar last updated on 09/Jul/23

((7−x )/( x−5))=(((7−x)/(x−5)))^3 ((x−5 )/( 7−x))=(((x−5)/(7−x)))^3 ↑↓ = ↑↓ x = 5

7−xx−5=(7−xx−5)3x−57−x=(x−57−x)3↑↓=↑↓x=5

Commented by Rasheed.Sindhi last updated on 09/Jul/23

Thanx Baliram!

ThanxBaliram!

Commented by BaliramKumar last updated on 09/Jul/23

Welcome Sir

WelcomeSir

Commented by Rasheed.Sindhi last updated on 09/Jul/23

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Answered by Frix last updated on 09/Jul/23

t=6−x ⇒ x=6−t ((((1+t))^(1/3) −((1−t))^(1/3) )/( ((1+t))^(1/3) +((1−t))^(1/3) ))=t (1−t)((1+t))^(1/3) =(1+t)((1−t))^(1/3) Both sides =0 if t=±1 ⇒ x=5∨x=7 Both sides =1 if t=0 ⇒ x=6 If it′s not obvious: (1−t)^3 (1+t)=(1+t)^3 (1−t) t_1 =−1 (1−t)^3 =(1+t)^2 (1−t) t_2 =1 (1−t)^2 =(1+t)^2 1−2t+t^2 =1+2t+t^2 −t=t t_3 =0

t=6−x⇒x=6−t1+t3−1−t31+t3+1−t3=t(1−t)1+t3=(1+t)1−t3Bothsides=0ift=±1⇒x=5∨x=7Bothsides=1ift=0⇒x=6Ifit′snotobvious:(1−t)3(1+t)=(1+t)3(1−t)t1=−1(1−t)3=(1+t)2(1−t)t2=1(1−t)2=(1+t)21−2t+t2=1+2t+t2−t=tt3=0

Answered by Rasheed.Sindhi last updated on 09/Jul/23

((((7−x))^(1/3) −((x−5))^(1/3) )/( ((7−x))^(1/3) +((x−5))^(1/3) ))−1=((6−x)/1)−1 ((−2((x−5))^(1/3) )/( ((7−x))^(1/3) +((x−5))^(1/3) ))=5−x ((2((x−5))^(1/3) )/( ((7−x))^(1/3) +((x−5))^(1/3) ))=x−5 ((7−x))^(1/3) +((x−5))^(1/3) =((2((x−5))^(1/3) )/(x−5))........(i) ((((7−x))^(1/3) −((x−5))^(1/3) )/( ((7−x))^(1/3) +((x−5))^(1/3) ))+1=((6−x)/1)+1 ((2((7−x))^(1/3) )/( ((7−x))^(1/3) +((x−5))^(1/3) ))=((7−x)/1) ((7−x))^(1/3) +((x−5))^(1/3) =((2((7−x))^(1/3) )/(7−x))......(ii) (i)&(ii): ((2((x−5))^(1/3) )/(x−5))=((2((7−x))^(1/3) )/(7−x)) (((x−5))^(1/3) /( ((7−x))^(1/3) ))=((x−5)/(7−x)) ((x−5)/( 7−x))=(((x−5)/(7−x)))^3 or ((7−x)/(x−5))=(((7−x)/(x−5)))^3 (((x−5)/(7−x)))^3 −((x−5)/( 7−x))=0 or (((7−x)/(x−5)))^3 −((7−x)/(x−5))=0 ((x−5)/( 7−x))((((x−5)/(7−x)))^2 −1)=0 or ((7−x)/(x−5))((((7−x)/(x−5)))^2 −1)=0 ((x−5)/( 7−x))=0 ∣ (((x−5)/(7−x)))^2 =1or ((7−x)/(x−5))=0 ∣ (((7−x)/(x−5)))^2 =1 ((x−5)/( 7−x))=0 ∣ ((7−x)/(x−5))=0 ∣ (((x−5)/(7−x)))^2 =1 x=5 ∣ x=7 ∣ x=6

7−x3−x−537−x3+x−53−1=6−x1−1−2x−537−x3+x−53=5−x2x−537−x3+x−53=x−57−x3+x−53=2x−53x−5........(i)7−x3−x−537−x3+x−53+1=6−x1+127−x37−x3+x−53=7−x17−x3+x−53=27−x37−x......(ii)(i)&(ii):2x−53x−5=27−x37−xx−537−x3=x−57−xx−57−x=(x−57−x)3or7−xx−5=(7−xx−5)3(x−57−x)3−x−57−x=0or(7−xx−5)3−7−xx−5=0x−57−x((x−57−x)2−1)=0or7−xx−5((7−xx−5)2−1)=0x−57−x=0∣(x−57−x)2=1or7−xx−5=0∣(7−xx−5)2=1x−57−x=0∣7−xx−5=0∣(x−57−x)2=1x=5∣x=7∣x=6

Answered by horsebrand11 last updated on 10/Jul/23

((−2((x−5))^(1/3) )/(2((7−x))^(1/3) )) = ((5−x)/(7−x)) (7−x)((x−5))^(1/3) = (x−5)((7−x))^(1/3) (7−x)^3 (x−5)=(x−5)^3 (7−x) (7−x)(x−5){(7−x)^2 −(x−5)^2 }=0 (7−x)(x−5)(2.(12−2x))=0

−2x−5327−x3=5−x7−x(7−x)x−53=(x−5)7−x3(7−x)3(x−5)=(x−5)3(7−x)(7−x)(x−5){(7−x)2−(x−5)2}=0(7−x)(x−5)(2.(12−2x))=0

Answered by Rasheed.Sindhi last updated on 10/Jul/23

((((7−x))^(1/3) −((x−5))^(1/3) )/( ((7−x))^(1/3) +((x−5))^(1/3) ))=6−x ((a−b)/(a+b))=6−x [ a=((7−x))^(1/3) , b=((x−5))^(1/3) ] ((a−b)/(a+b))∙((a^2 +ab+b^2 )/(a^2 −ab+b^2 ))∙((a^2 −ab+b^2 )/(a^2 +ab+b^2 ))=6−x ((a^3 −b^3 )/(a^3 +b^3 ))∙((a^2 −ab+b^2 )/(a^2 +ab+b^2 ))=6−x (((7−x)−(x−5))/((7−x)+(x−5)))∙((a^2 −ab+b^2 )/(a^2 +ab+b^2 ))=6−x ((2(6−x))/2)∙((a^2 −ab+b^2 )/(a^2 +ab+b^2 ))=6−x (6−x)∙((a^2 −ab+b^2 )/(a^2 +ab+b^2 ))−(6−x)=0 (6−x)(((a^2 −ab+b^2 )/(a^2 +ab+b^2 ))−1)=0 6−x=0 ∣ ((a^2 −ab+b^2 )/(a^2 +ab+b^2 ))−1=0 x=6✓ ∣ ((a^2 −ab+b^2 )/(a^2 +ab+b^2 ))=1 ∣ a^2 −ab+b^2 =a^2 +ab+b^2 ∣ ab=0 ∣ a=0 ∣ b=0 ∣ ((7−x))^(1/3) =0 ∣ ((x−5))^(1/3) =0 ∣ x=7✓ ∣ x=5✓

7−x3−x−537−x3+x−53=6−xa−ba+b=6−x[a=7−x3,b=x−53]a−ba+b⋅a2+ab+b2a2−ab+b2⋅a2−ab+b2a2+ab+b2=6−xa3−b3a3+b3⋅a2−ab+b2a2+ab+b2=6−x(7−x)−(x−5)(7−x)+(x−5)⋅a2−ab+b2a2+ab+b2=6−x2(6−x)2⋅a2−ab+b2a2+ab+b2=6−x(6−x)⋅a2−ab+b2a2+ab+b2−(6−x)=0(6−x)(a2−ab+b2a2+ab+b2−1)=06−x=0∣a2−ab+b2a2+ab+b2−1=0x=6✓∣a2−ab+b2a2+ab+b2=1∣a2−ab+b2=a2+ab+b2∣ab=0∣a=0∣b=0∣7−x3=0∣x−53=0∣x=7✓∣x=5✓

Answered by Rasheed.Sindhi last updated on 10/Jul/23

((((7−x))^(1/3) −((x−5))^(1/3) )/( ((7−x))^(1/3) +((x−5))^(1/3) ))=6−x=(((7−x)−(x−5))/((7−x)+(x−5))) ((1−(((x−5))^(1/3) /( ((7−x))^(1/3) )))/(1+(((x−5))^(1/3) /( ((7−x))^(1/3) ))))=((1−((x−5)/(7−x)))/(1+((x−5)/(7−x)))) [x≠7] (((x−5))^(1/3) /( ((7−x))^(1/3) ))=((x−5)/(7−x)) ((x−5)/(7−x))=(((x−5)/(7−x)))^3 x=5,6................(i) Similarly, (((((7−x))^(1/3) /( ((x−5))^(1/3) ))−1)/((((7−x))^(1/3) /( ((x−5))^(1/3) ))+1))=((((7−x)/(x−5))−1)/(((7−x)/(x−5))−1)) [x≠5] (((7−x))^(1/3) /( ((x−5))^(1/3) ))=((7−x)/(x−5)) ((7−x)/(x−5))=(((7−x)/(x−5)))^3 x=6,7...............(ii) From (i) & (ii): x=5 or 6 or 7

7−x3−x−537−x3+x−53=6−x=(7−x)−(x−5)(7−x)+(x−5)1−x−537−x31+x−537−x3=1−x−57−x1+x−57−x[x≠7]x−537−x3=x−57−xx−57−x=(x−57−x)3x=5,6................(i)Similarly,7−x3x−53−17−x3x−53+1=7−xx−5−17−xx−5−1[x≠5]7−x3x−53=7−xx−57−xx−5=(7−xx−5)3x=6,7...............(ii)From(i)&(ii):x=5or6or7

Commented by Frix last updated on 10/Jul/23

From ((...)/(7−x)) ⇒ x≠7

From...7−x⇒x≠7

Commented by Rasheed.Sindhi last updated on 10/Jul/23

Yes sir, you′re very right! I′ve modified my answer.

Yessir,you′reveryright!I′vemodifiedmyanswer.

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