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If a + b + c = 7 and a3+b3+c33abc=175a^3 + b^3 + c^3-3abc = 175, then what is the value of (ab + bc + ca)?

A8

B7

C6

D9

Answer:

A. 8

Read Explanation:

Solution:

Given:

a + b + c = 7

a3 + b3 + c3 - 3abc = 175

Concept used:

a+ b3 + c3 - 3abc = (a + b + c)[(a + b + c)2 - 3(ab + bc + ca)]

Calculation:

a3+b3+c33abc=(a+b+c)[(a+b+c)23(ab+bc+ca)]a^3+b^3+c^3-3abc=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]

175=7×[(7)23(ab+bc+ca)]175=7\times{[(7)^2-3(ab+bc+ca)]}

25=493(ab+bc+ca)25=49-3(ab+bc+ca)

⇒ 24 = 3(ab + bc + ca)

⇒ ab + bc + ca = 8

∴ The value of given identities is 8.


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ഒരു സംഖ്യയുടെ 4 മടങ്ങിനെക്കാൾ 5 കുറവ്, ആ സംഖ്യയുടെ 3 മടങ്ങിനെക്കാൾ 3 കൂടുതലാണ്. എന്നാൽ സംഖ്യ ഏത് ?