The expression x3(x−y)7−y3(x−y)7 can be rewritten as (x−y)m(x2+y2+nxy). What is the value of m ?

The value of m is 8

The expression is given as:

Factor out (x - y)^7 in the above expression

Expand out (x^3 - y^3) as the difference of two cubes

Combine the common factors

Rewrite the expression as:

The expression format is given as:

So, we have:

By comparison, we have:

Hence, the value of m is 8

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Step-by-step explanation:

Given the expression:

x³(x−y)⁷−y³(x−y)⁷

This can also be written as (x³-y³)(x-y)⁷

(x³-y³)(x-y)⁷=  (x-y)⁷ (x³-y³)

Expand  (x³-y³)

x³-y³ = (x-y)(x²+xy+y²)

Substitute back into the original expression:

(x-y)⁷ (x³-y³) = (x-y)⁷(x-y)(x²+xy+y²)

(x-y)⁷ (x³-y³) = (x-y)⁸(x²+xy+y²)

(x-y)⁷ (x³-y³) = (x-y)⁸(x²+y²+xy)

Comparing the result with (x−y)^m(x^2+y^2+nxy).

(x-y)⁸ = (x-y)^m

m = 8

Also:

nxy = xy

n = xy/xy

n = 1

Hence the value of m is 8

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.