Difference between revisions of "1975 AHSME Problems/Problem 27"
Megaboy6679 (talk | contribs) (→Solution 1) |
Megaboy6679 (talk | contribs) (→Solution 1) |
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<cmath>p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6.</cmath> | <cmath>p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6.</cmath> | ||
− | By Vieta's formulas, <math>p + q + r = 1</math>, <math>pq + pr + qr = 1</math>, and <math>pqr = 2</math>. Squaring the equation <math>p + q + r = 1</math>, we get | + | By [[Vieta's formulas]], <math>p + q + r = 1</math>, <math>pq + pr + qr = 1</math>, and <math>pqr = 2</math>. Squaring the equation <math>p + q + r = 1</math>, we get |
<cmath>p^2 + q^2 + r^2 + 2pq + 2pr + 2qr = 1.</cmath> | <cmath>p^2 + q^2 + r^2 + 2pq + 2pr + 2qr = 1.</cmath> | ||
Subtracting <math>2pq + 2pr + 2qr = 2</math>, we get | Subtracting <math>2pq + 2pr + 2qr = 2</math>, we get |
Revision as of 00:06, 19 March 2023
Problem
If and
are distinct roots of
, then
equals
Solution 1
If is a root of
, then
, or
Similarly,
, and
, so
By Vieta's formulas, ,
, and
. Squaring the equation
, we get
Subtracting
, we get
Therefore, . The answer is
.
Solution 2(Faster)
We know that . By Vieta's formulas,
,
, and
.
So if we can find
, we are done. Notice that
, so
, which means that
~pfalcon