You would have to determine if<span> the following </span>lengths<span> make an </span>acute<span>, right or </span>obtuse triangle<span>. Plug in each set of </span>lengths<span> into the Pythagorean Theorem.</span>
Answer: <em>0</em>
Step-by-step explanation:
<em>Simply the answer is </em><em>0</em><em> because anything multiplied by </em><em>0</em><em> will result in </em><em>0</em>
<em>4x(0) =</em><em> 0 </em>
the correct answer would be -24x because 9+12 is 21 and 21 - 45 would be -24
<u>1</u><u>.</u><u> </u><u>(ab + bc) (ab – bc) + (bc + ca) (bc – ca) + (ca + ab) (ca – ab) = 0</u>
We know that (a+b)(a-b) = a²-b²
(ab + bc)(ab -bc) can be written as a²b² - b²c²
(bc + ca)(bc -ca) can be written as b²c² - c²a²
(ca + ab)(ca - ab) can be written as c²a² - a²b²
→ a²b² - b²c² + b²c² - c²a² + c²a² - a²b²
→ a²b² - a²b² - b²c² + b²c² - c²a² + c²a²
→ 0
<u>2</u><u>.</u><u> </u><u>(a + b + c) (a² + b² + c² – ab – bc – ca) = a³ + b³+ c³ – 3abc</u>
→ a³ + ab² + ac² -a²b - abc -ca² + a²b + b³ + bc² - ab² - b²c - abc + a²c + b²c + c³ - abc - bc² - c²a
→ a³ + b³+ c³ + (- abc - abc - abc) + (ab² - ab² )+ (ac² - ca² ) -(a²b + a²b )+ (bc² - bc² )+ (a²c - c²a) + (b²c - b²c)
→ a³+b³+c³ - 3 abc .
<u>3</u><u>.</u><u> </u><u>(p – q) (p² + pq + q²) = p³ – q³. </u>
→ p³ + p²q + pq² - p²q - pq² - q³
→ p³ - q³ +(p²q - p²q) + (pq² - pq²)
→ p³ - q³