Q1.
We already know that x² − y² = (x − y)(x + y). Further, we have verified that x³ − y³ = (x − y)(x² + xy + y²). Observe that x − y is a common factor of x² − y² and x³ − y³. Do you think x − y is also a factor of x⁴ − y⁴? Note that x⁴ − y⁴ = (x²)² − (y²)² = (x² − y²)(x² + y²). Can you see how x − y is a factor of x⁴ − y⁴?
Answer
Yes. The hint does almost all the work — the first bracket already contains x − y.
x⁴ − y⁴ = (x²)² − (y²)²
= (x² − y²)(x² + y²) — difference of two squares
= (x − y)(x + y)(x² + y²) — difference of two squares again
So x⁴ − y⁴ = (x − y)(x + y)(x² + y²), and x − y is a factor.
= (x² − y²)(x² + y²) — difference of two squares
= (x − y)(x + y)(x² + y²) — difference of two squares again
So x⁴ − y⁴ = (x − y)(x + y)(x² + y²), and x − y is a factor.
Equally, the direct division gives
x⁴ − y⁴ = (x − y)(x³ + x²y + xy² + y³)
and the two answers agree, since (x + y)(x² + y²) = x³ + x²y + xy² + y³.
Why x − y must be a factor, without any algebra: put x = y in the expression x⁴ − y⁴. You get 0. Any expression that vanishes whenever x = y has x − y as a factor — that is the same reasoning that made x − y a factor of x² − y² and of x³ − y³. You will meet this as the Factor Theorem in Grade 10.