The expression \( a = \sqrt[3]{7+5\sqrt{2}} + \sqrt[3]{7-5\sqrt{2}} \) is proven to be an integer, specifically equal to 2. This conclusion is reached by expanding \( (1 \pm \sqrt{2})^3 \) using the binomial theorem, which simplifies to \( 7 \pm 5\sqrt{2} \). The calculations confirm that \( \sqrt[3]{7 + 5\sqrt{2}} + \sqrt[3]{7 - 5\sqrt{2}} \) results in \( (1 + \sqrt{2}) + (1 - \sqrt{2}) = 2 \).
PREREQUISITESMathematics students, educators, and enthusiasts interested in algebraic identities and integer solutions involving roots.
lfdahl said:Let $a = \sqrt[3]{7+5\sqrt{2}} + \sqrt[3]{7-5\sqrt{2}}$. Show (without the use of a calculator), that $a$ is an integer.
My solution:lfdahl said:Let $a = \sqrt[3]{7+5\sqrt{2}} + \sqrt[3]{7-5\sqrt{2}}$. Show (without the use of a calculator), that $a$ is an integer.
kaliprasad said:$a = \sqrt[3]{7+5\sqrt{2}} + \sqrt[3]{7-5\sqrt{2}}$.
we have a is real
cube both sides to get using $(x+y)^3= x^3+y^3 + 3xy(x+y)$
$a^3= 7+5\sqrt{2} + 7-5\sqrt{2}+ 3 (\sqrt[3]{7+5\sqrt{2}})(\sqrt[3]{7-5\sqrt{2}})a$
$= 14 + 3a$
hence $a^3+3a-14=0$
by rational root theorem 2 is a root
hence $a^3-4a + 7a-14 =0$
or $(a-2)(a^2 + 2a) +7(a- 2) = (a-2)(a^2+2a+7)=0$
so a = 2 or $a^2+2a+7=0$ which gives complex roots
so a = 2 as a is real
now I leave some one to prove that 2 is integer
This is a suprisingly short and elegant way to "crack the nut". Thankyou, Opalg, for your participation!Opalg said:My solution:
[sp]Expand $(1 \pm \sqrt2)^3$ binomially to get $$(1 \pm \sqrt2)^3 = 1 \pm3\sqrt2 + 6 \pm2\sqrt2 = 7 \pm 5\sqrt2.$$ So $1 \pm \sqrt2 = \sqrt[3]{7\pm 5\sqrt2}$, and $\sqrt[3]{7 + 5\sqrt2} + \sqrt[3]{7 - 5\sqrt2} = (1 + \sqrt2) + (1 - \sqrt2) = 2.$
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