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  • FACTORIZE WITHOUT CALCULATING THE CUBES :
  • (1/2x - 1/3y)^3 + (1/3y - 1/4z)^3 + (1/4z - 1/2x)^3
  • (a - 2b)^3 + (2b - 4c)^3 + (4c - a)^3
  • FIND :
  • 8x^3 + y^3 - 30xy + 125 if 2x + y = -5
  • x^3 - 27y^3 + 27xy + 27 if x = 3y - 3
  • FACTORIZE THE FOLLOWING :
  • 289a^2 - 529b^2
  • 7root2x^2 - 10x - 4root2
  • 64x^3 + 8
  • x^2 - x - 110
  • a^3 + 3root3b^3
  • ONE OF THE FACTORS OF 4x^3 + 2x^2 - 16x + 10 IS 2x + 5 FIND THE OTHER FACTORS AND HENCE ALL THE FACTORS

PLS EXPERTS I REQUEST U TO SOLVE THIS AS FAST AS POSSIBLE BECAUSE THIS IS AN ASSIGNMENT THAT I HAVE TO SUBMIT TOMORROW

AND I WANT THE STEPS PLS BECAUSE WHEN YOU SUGGEST SOME PAGES TO REFER I DONT UNDERSTAND

I

1Let A = 12x - 13y ; B = 13y - 14z ; C = 14z - 12xNow, A + B + C = 12x - 13y + 13y - 14z + 14z - 12x = 0Since, A + B + C = 0,  SoA3 + B3 + C3 = 3 ABC⇒ 12x - 13y3 + 13y - 14z3 + 14z - 12x3 = 3 12x - 13y13y - 14z14z - 12x2Let x = a - 2b ; y = 2b - 4c ; z = 4c - a Now, x + y + z = a - 2b + 2b - 4c + 4c - a = 0Since,  x + y + z = 0, thenx3 + y3 + z3 = 3 xyz⇒ a - 2b3 + 2b - 4c3 + 4c - a3 = 3 a - 2b2b - 4c4c - a 

3i.  We have, 2x + y = -5Now, 8x3 + y3 - 30 xy + 125= 2x3 + y3 + 53 - 3 2x y 5= 2x + y + 5 2x2 + y2 + 52 - 2x y - y 5 - 5 2x= 2x + y + 5 4x2 + y2 + 25 - 2 xy - 5y - 10x= -5 + 5  4x2 + y2 + 25 - 2 xy - 5y - 10x= 0ii.  We have, x = 3y - 3 ⇒ x - 3y = -3Now, x3 - 27y3 + 27 + 27 xy= x3 + -3y3 + 33 - 3 x -3y 3= x - 3y + 3 x2 + -3y2 + 32 - x -3y - -3y 3 - 3 x= x - 3y + 3 x2 + 9y2 + 9 + 3 xy + 9y - 3x= -3 + 3  x2 + 9y2 + 9 + 3 xy + 9y - 3x= 0

4.a. 289 a2 - 529 b2   = 17a2 - 23b2     x2 - y2 = x + y x - y   = 17a + 23b 17a - 3bb. 72x2 - 10x - 42  =72x2 - 14x + 4x - 42  =72xx - 2 + 4x - 2  =72x + 4x - 2c. 64x3 + 8  =4x3 + 23  =4x + 24x2 + 22 - 4x2  =4x + 216x2 + 4 - 8x  =82x+14x2 + 1 - 2xd. x2 - x - 110    =x2 - 11x + 10x -110    =xx - 11 + 10x - 11    =x + 10x - 11e. a3 + 33b3   =a3 + 33   =a+3a2 + 3b2 - a3b   =a + 3ba2 +3b2 -3ab




 

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a^3 + b^3 +c^3 -3abc = (a+b+c)(a^2+b^2+c^2+ab+bc+ac)

here a=1/2x-1/3y and b=1/3y-1/4z and c=1/4z-1/2x

so

a+b+c = 1/2x -1/3y +1/3y -1/4z +1/4z -1/2x

a+b+c =0

so

a^3 + b^3 + c^3 -3abc =0

a^3 +b^3 +c^3 =3abc

a^3 +b^3 +c^3 = 3 (1/2x -1/3y)(1/3y -1/4z)(1/4z - 1/2x)

Now you got to calculate that

Q.2) Here a +b +c = a -2b +2b -4c +4c -a

a+b+c =0

so

a^3 +b^3 +c^3 = 3abc

a^3 +b^3 +c^3 = 3(a-2b)(2b-4c)(4c-a)

Now calculate that

Plzz thumbs up!! :D

  • 1

can u pls calculate?? plshh

  • 0

and wat Abt factorize ques?

  • 0
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