tag:blogger.com,1999:blog-2400513859305780710.post9209059751770434335..comments2019-07-11T22:41:53.239-07:00Comments on Mathematical Food For Thought: Sumsine. Topic: Trigonometry/Geometry. Level: AMC.Jeffrey Wanghttp://www.blogger.com/profile/11114458640271201663noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-2400513859305780710.post-88410648351358263592007-04-01T13:03:59.000-07:002007-04-01T13:03:59.000-07:00Zeroes are $a-d,a,a+d$$3a=\frac{4}{7}$, and we are...Zeroes are $a-d,a,a+d$<br><br>$3a=\frac{4}{7}$, and we are trying to find $a^3-ad^2$<br><br>We&#39;re also given $a(a-d)+a(a+d)+(a-d)(a+d)=3a^2-d^2=0$ so $3a^2=d^2$<br><br>Then we want $a^3-3a^3=-2a^3$ but we already know $a$...the answer is $\frac{(-2)(4^3)}{21^3}$ whatever that is...?...is this correct?ch1n353ch3s54a1lnoreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-42588029267878621532007-04-02T08:55:10.000-07:002007-04-02T08:55:10.000-07:00Close. The product of the roots is actually $-\fr...Close. The product of the roots is actually$ -\frac{K}{7} $, so you need to multiply by a factor of$ -7 \$.paladin8noreply@blogger.com