(If a³+ā³ = 50√7)
So prove that a = √6 + √7
(Here ā³ means a to the power minus 3)
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Step-by-step explanation:
Given 0<50−−√−7<0.1 we have that, using y=50−−√ :
0<y3–3∗7∗y2+3∗72∗y−73<0.001 (a)
where we raised all sides of the inequality to the third power.
Now, we are interested in showing (50−−√+7)3 differs from an integer less than 0.001. Again using y=50−−√ we get for the above
(y+7)3=y3+3∗7∗y2+3∗72∗y+73 (b)
Now subtract a from b: 6∗7∗y2+2∗73 . Replacing y we get that
b−a=6∗7∗50+2∗73 , an integer.
Since a<0.001 then the difference between b and an integer is less than 0.001
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