11 Simplify the following \( 111 \quad(m-2 n)\left(m^{2}-6 m n-n^{2}\right) \) \( 11.2 \frac{x^{2}-4}{x^{2}-5 x+6} \times \frac{3-x}{x+2} \) \( 1.3 \quad \frac{y^{3}-1}{y^{2}+y+1}-\frac{4 y^{2}-3 y-1}{4 y+1} \)
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Bonus Knowledge
To simplify the expression \(111(m-2n)(m^{2}-6mn-n^{2})\), we first need to factor \(m^{2}-6mn-n^{2}\). This expression can be factored as \((m-n)(m-5n)\). Now substituting back into the expression gives us \(111(m-2n)(m-n)(m-5n)\). This is a factored form, ready for any further operations. For the second expression \( \frac{x^{2}-4}{x^{2}-5x+6} \times \frac{3-x}{x+2} \), we can simplify \(x^{2}-4\) to \((x-2)(x+2)\) and \(x^{2}-5x+6\) to \((x-2)(x-3)\). Thus, we have: \[ \frac{(x-2)(x+2)}{(x-2)(x-3)} \times \frac{3-x}{x+2} \] The \( (x-2) \) terms cancel out, leaving us with: \[ \frac{(3-x)}{(x-3)} \] And since \(3-x = -(x-3)\), the final simplified expression is \(-1\). For the last expression \( \frac{y^{3}-1}{y^{2}+y+1}-\frac{4y^{2}-3y-1}{4y+1} \), we can factor \(y^{3}-1\) to \((y-1)(y^{2}+y+1)\). Substituting yields: \[ \frac{(y-1)(y^{2}+y+1)}{y^{2}+y+1} - \frac{4y^{2}-3y-1}{4y+1} \] This cancels out the common term \(y^{2}+y+1\), simplifying it to \(y-1 - \frac{4y^{2}-3y-1}{4y+1}\). To combine these fractions, we find a common denominator: \[ (y-1)(4y+1) - (4y^{2}-3y-1) \] Expanding and rearranging gives us a combined expression which can then be simplified further by collecting like terms.