Question

The remainder when x^4-y^4 is divided by x+y is​

Answer 1

$= > \tt {x}^{4} - {y}^{4} \\ \\ = > \sf {( {x}^{2} )}^{2} - {( {y}^{2} )}^{2} \\ \\ = > \tt \{{x}^{2} - {y}^{2} \} \{ {x}^{2} + {y}^{2} \} \\ = > \bf \{ {(x)}^{2} - {(y)}^{2} \}\{ {x}^{2} + {y}^{2} \} \\ \sf = > \green{ \{(x + y)(x - y) \} \{ {x}^{2} + {y}^{2} \}}$

Now,

$= > \sf \frac{\{(x + y)(x - y) \} \{ {x}^{2} + {y}^{2} \}}{x + y} \\ \\ \sf= > \purple{(x - y)( {x}^{2} + {y}^{2} )}$

Since,

• x⁴ - y⁴ gets completely divided by x+y, no remainder is left(here).
• Thus, Remainder on dividing x⁴ - y⁴ by x + y is 0