(a + b)³-(a - b)³- 6a²b

Mathematics

2 answers:

maksim [4K]3 years ago

7 0

Step-by-step explanation:

$(a + b)^{3} - (a - b)^{3} - 6 {a}^{2} b \\ a ^{3} + 3a ^{2} b + 3ab ^{2} + b ^{3} - (a ^{3} - 3a ^{2} b + 3ab ^{2} - b ^{3}) - 6 {a}^{2} b \\ a ^{3} + 3a ^{2} b + 3ab ^{2} + b ^{3} - a ^{3} + 3a ^{2} b - 3ab ^{2} + b ^{3}) - 6 {a}^{2} b \\ 3a ^{2} b + b ^{3} + 3a ^{2} b + b ^{3} - 6 {a}^{2} b \\ 6 {a}^{2} b + 2b ^{3} - 6 {a}^{2} b \\ 2b ^{3}$

Andru [333]3 years ago

3 0

Answer:

2b^3

Step-by-step explanation:

$(a+b)^3=a^3+b^3+3ab(a+b)\\(a-b)^3=a^3-b^3-3ab(a-b)\\(a+b)^3-(a-b)^3-6a^2b\\=a^3+b^3+3ab(a+b)-[a^3-b^3-3ab(a-b)]-6a^2b\\=2b^3$

You might be interested in

The length of a photograph is 1 cm less than twice the width. The area is 91cm^2. Find the width and length.

Brums [2.3K]

Let the width = x
=>length = 2x-1

91= x(2x-1)=2x^2-x
2x^2-x-91=0
=(2x+13)(x-7)
=> x=-13/2=> no
=>x= 7 cm

7 0

3 years ago

Make x the subject in the relation

tatyana61 [14]

Cant read your writing-

7 0

3 years ago

Define f(0,0) in a way that extends f to be continuous at the origin. f(x, y) = ln ( 19x^2 - x^2y^2 + 19 y^2/ x^2 + y^2) Let f (

kirill115 [55]

Answer:

f(0,0)=ln19

Step-by-step explanation:

$f(x,y)=ln(\frac{19x^2-x^2y^2+19y^2}{x^2+y^2})$ is given as continuous function, so there exist $lim_{(x,y)\rightarrow(0,0)}f(x,y)$ and it is equal to f(0,0).

Put x=rcosA annd y=rsinA

$f(r,A)=ln(\frac{19r^2cos^2A-r^2cos^2A*r^2sin^2A+19r^2sin^2A}{r^cos^2A+r^2sin^2A})=ln(\frac{19r^2(cos^2A+sin^2A)-r^4cos^2Asin^a}{r^2(cos^2A+sin^2A)})$