The discussion centers around demonstrating that the expression \( (7+5\sqrt{2})^{1/3} + (7-5\sqrt{2})^{1/3} \) is an integer, exploring the mathematical reasoning behind this claim without the use of a calculator.
There appears to be agreement on the method used to demonstrate that \( a \) is an integer, as one participant confirms the solution presented by another. However, the discussion does not explore alternative methods or challenge the reasoning provided.
The discussion relies on the assumption that the binomial expansion is correctly applied and that the identities used are valid without further verification.
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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