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2 years ago

(w-w^2)^4=?.......................

Mathematics

1 answer:

Setler [38]2 years ago

3 0

Answer:

w^8 - 4w^7 + 6w^6 - 4w^5 + w^4 .

Step-by-step explanation:

(w - w^2)^4

= w^4 ( 1 - w)^4

By the binomial theorem:

= w^4( 1 - 4w + 6w^2 - 4w^3 + w^4)

= w^4 - 4w^5 + 6w^6 - 4w^7 + w^8

= w^8 - 4w^7 + 6w^6 - 4w^5 + w^4 ( Standard form).

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2 years ago

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3 years ago

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Answer:

Step-by-step explanation:

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2 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

Can you guys help we with this question.

satela [25.4K]

Answer:

Step-by-step explanation:

(2x+1/(2x))^5 *(2x -1/(2x))^5

= ((2x)^2 -1/(2x)^2)^5 (a+b)*(a-b) =a2-b2

= (4x^2-1/4(x)^2)^5

now

x =4x^2. ,a = 1/4(x)^2 ,n =5

we have

general term = Cr *x^r *a^(n-r)

= Cr * (4x^2)^r * (1/4(x)^2)^(n-r)

= Cr *4^r * X^2r * 1/( 4^(n-r) *x^(2n-2r)

= Cr * 4^r/4^(n-r) * x^(2r)/x^(2n-2r)

= Cr * 4(2r-n) *x(4r-2n)

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4r -10=2

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r = 3

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C(5,3) * 4^(2*3-5)

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4 0

2 years ago

Read 2 more answers

(w-w^2)^4=?.......................​

(w-w^2)^4=?.......................