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Factor completely.

1-x^(6)y^(4)
Answer:

Factor completely.\newline1x6y4 1-x^{6} y^{4} \newlineAnswer:

Full solution

Q. Factor completely.\newline1x6y4 1-x^{6} y^{4} \newlineAnswer:
  1. Recognize squares: Recognize the expression as a difference of squares.\newlineThe given expression is 1x6y41 - x^{6}y^{4}. We can see that both 11 and x6y4x^{6}y^{4} are perfect squares because 11 is the square of 11 and x6y4x^{6}y^{4} is the square of x3y2x^{3}y^{2}.
  2. Apply formula: Apply the difference of squares formula.\newlineThe difference of squares formula is a2b2=(ab)(a+b)a^2 - b^2 = (a - b)(a + b). Here, a2a^2 is 11 and b2b^2 is x6y4x^{6}y^{4}, so aa is 11 and bb is x3y2x^{3}y^{2}.
  3. Factor expression: Factor the expression using the difference of squares formula.\newlineUsing the values of aa and bb from Step 22, we get:\newline1x6y4=(1x3y2)(1+x3y2)1 - x^{6}y^{4} = (1 - x^{3}y^{2})(1 + x^{3}y^{2})
  4. Check further factorization: Check for any further factorization.\newlineBoth factors (1x3y2)(1 - x^{3}y^{2}) and (1+x3y2)(1 + x^{3}y^{2}) cannot be factored further using real numbers, as they are not differences or sums of squares or any other factorable form.

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