x³+y³+z³=k thereeeeeeeeeeee
Answers
We understand x , y and z as integers belonging to IZ => negative and positive Natural integers.
x3+y3+z3=k with k is integer from 1 to 100
solution x=0 , y=0 and z=1 and k= 1
For K= 1 , we have the following solutions (x,y,x) = (1,0,0) ; or (0,1,0) ; or (0,0,1) ,
For k =1 also (9,-8,-6) or (9,-6,-8) or (-8,-6,9) or (-8,9,-6) or (-6,-8,9) or (-6,9,8)
And (-1,1,1) or (1,-1,1)
=>(x+y)3−3x2−3xy2+z3=k
=>(x+y+z)3−3(x+y)2.z−3(x+y).z2=k
=>(x+y+z)3−3(x+y)z[(x+y)−3z]=k
let y=α and z=β
=>x3=−α3−β3+k
For k= 2 we have (x,y,z) = (1,1,0) or (1,0,1) or (0,1,1)
Also for (x,y,z) = (7,-6,-5) or (7,-6,-5) or (-6,-5,7) or (-6,7,-5) or (-5,-6,7) or (-5,7,-6)
For k= 3 we have 1 solution : (x,y,z) = (1,1,1)
For k= 10 , we have the solutions (x,y,z) = (1,1,2) or (1,2,1) or (2,1,1)
For k= 9 we have the solutions (x,y,z) = (1,0,2) or (1,2,0) or (0,1,2) or (0,2,1) or (2,0,1) or (2,1,0)
For k= 8 we have (x,y,z) = ( 0,0,2) or (2,0,0) or (0,2,0)
For k= 17 => (x,y,z) = (1,2,2) or (2,1,2) or ( 2,2,1)
For k = 24 we have (x,y,z) = (2,2,2)
For k= 27 => (x,y,z) = (0,0,3) or (3,0,0) or (0,3,0)
for k= 28 => (x,y,z) = (1,0,3) or (1,3,0) or (1,3,0) or (1,0,3) or (3,0,1) or (3,1,0)
For k=29 => (x,y,z) = (1,1,3) or (1,3,1) or (3,1,1)
For k = 35 we have (x,y,z) = (0,2,3) or (0,3,2) or (3,0,2) or (3,2,0) or 2,0,3) or (2,3,0)
For k =36
we have also solution : x=1,y=2andz=3=>
13+23+33=1+8+27=36 with k= 36 , we have the following
we Have : (x, y,z) = (1, 2, 3) ; (3,2,1); (1,3,2) ; (2,1,3) ; (2,3,1), and (3,1,2)
For k= 43 we have (x,y,z) = (2,2,3) or (2,3,2) or (3,2,2)
For k = 44 we have ( 8,-7,-5) or (8,-5,-7) or (-5,-7,8) or ( -5,8,-7) or (-7,-5,8) or (-7,8,-5)
For k =54 => (x,y,z) = (13,-11,-7) ,
for k = 55 => (x,y,z) = (1,3,3) or (3,1,3) or (3,1,1)
and (x,y,z) = (10,-9,-6) or (10,-6,-9) or ( -6,10,-9) or (-6,-9,10) or (-9,10,-6) or (-9,-6,10)
For k = 62 => (x,y,z) = (3,3,2) or (2,3,3) or (3,2,3)
For k =64 => (x,y,z) = (0,0,4) or (0,4,0) or (4,0,0)
For k= 65 => (x,y,z) = (1,0,4) or (1,4,0) or (0,1,4) or (0,4,1) or (4,1,0) or (4,0,1)
For k= 66 => (x,y,z) = (1,1,4) or (1,4,1) or (4,1,1)
For k = 73 => (x,y,z) = (1,2,4) or (1,4,2) or (2,1,4) or (2,4,1) or (4,1,2) or (4,2,1)
For k= 80=> (x,y,z)= (2,2,4) or (2,4,2) or (4,2,2)
For k = 81 => (x,y,z) = (3,3,3)
For k = 90 => (x,y,z) = (11,-9,-6) or (11,-6,-9) or (-9,11,-6) or (-9,-6,11) or (-6,-9,11) or (-6,11,-9)
k = 99 => (x,y,z) = (4,3,2) or (4,2,3) or (2,3,4) or (2,4,3) or ( 3,2,4 ) or (3,4,2)
(x,y,z) = (5,-3,1) or (5,1,-3) or (-3,5,1) or (-3,1,5) or (1,-3,5) or (1,5,-3)
=> 5^3 + (-3)^3 +1 = 125 -27 +1 = 99 => for k = 99
For K = 92
6^3 + (-5)^3 +1 = 216 -125
+1 = 92
8^3 +(-7)^3
Please mark me as brainlist....
Answer:
Step-by-step explanation:
ANSWER:-
Answering according to Attachments provided.
In attachment 1,
1. \sf{3 \times 4^3}3×43
So, we know that 4^3 is actually 4*4*4 = 64
So, indirectly it is 3*64 = 192 is the first answer.
2. Next is \sf{4x^3+2x^3}4x3+2x3
4 and 2 will get multiplied, and exponents will get added.
\sf{(4 \times 2) ^{(3+3)}}(4×2)(3+3)
3. Anything to the power 0 is 1. So the answer is 1.
4. The question is done in the attachment.
Multiplying Polynomials Attachment:-
6y(2y+3)
\sf{ \mapsto 12y^2 + 18y}↦12y2+18y
(2a-1)(8a-5)
= \sf{16a^2 - 10a - 8a + 5}16a2−10a−8a+5
(4p-1)^2
So, we apply an identity for this!
\sf{(a-b)^2 = a^2 + b^2 - 2ab}(a−b)2=a2+b2−2ab
\sf{(4p - 1)^2 = (4p)^2 + (1)^2 - 2(4p \times 1)}(4p−1)2=(4p)2+(1)2−2(4p×1)
\sf{ \implies 16p^2 + 1 - 8p}⟹16p2+1−8p
(2n+2)(6n+1)
\sf{12n^2 + 2n + 12n + 2}12n2+2n+12n+2
\sf{ \longrightarrow 12n^2 + 14n + 2}⟶12n2+14n+2
(7x-6)(5x+6)
\sf{35x^2 + 42x - 30x -12}35x2+42x−30x−12
\sf{35x^2 + 12x -12}35x2+12x−12
Dividing the Polynomials Attachment:-
This is to be done in detail, so let's see!
1. \sf{ - \dfrac{9a^3b^2}{3ab}}−3ab9a3b2
9 and 3 cancelld, ab cancelled with numerator and the answer is -3a^2b.
2. Taking 9 and r common from the numerator,
we can get the numerator as 9r(2r^4+4r^3+3r^2)
Now, we can cancel off 9r, and the required answer is :-
\sf{2r^4+4r^3+3r^2}2r4+4r3+3r2
3. Done in the attachment, kindly refer to it.
4th one, the coefficients are reduced to 15x^3+ x^2+15x/5r^2 and no change in the alphabet portion, surely r and x can never be canclled.
5. we can not perform this operation as their is no common factor between the both operation.