Question
Download Solution PDFजर (a + b - c) = 20, आणि a2+ b2 + c2 = 152 असल्यास, a3 + b3 - c3 + 3abc चे मूल्य शोधा.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिलेले आहे:
(a + b - c) = 20;
a2+ b2 + c2 = 152
वापरलेले सूत्र:
a3 + b3 = (a + b)3 - 3ab × (a + b)
गणना:
आपल्याला, सदर समीकरणात 3 चल आणि 2 समीकरणे दिलेली आहेत.
तर, समजा, c = 0
⇒ (a + b) = 20
⇒ a2 + b2 = 152
आता,
(a + b) = 20
दोन्ही बाजूंचा वर्ग केल्यास,
⇒ a2 + b2 + 2ab = 400
⇒ 152 + 2ab = 400
⇒ 2ab = 400 - 152 = 248
⇒ ab = 248/2 = 124
a3 + b3 = (a + b)3 - 3ab × (a + b)
⇒ (20)3 - 3 × 124 × 20
⇒ 8000 - 7440 = 560
∴ 560 हे योग्य उत्तर आहे.
Alternate Method संकल्पना:
आवश्यक पदावली शोधण्यासाठी बीजगणितीय समरूपता वापरू.
गणना:
घनांच्या बेरजेसाठीची समरूपता वापरू:
आपल्याला माहीत आहे की,
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
जर, c ⇒ - c, तर
a3 + b3 - c3 + 3abc = (a + b - c){(a2 + b2 + c2 - ab - b(-c) - (-c)a)}
⇒ a3 + b3 - c3 + 3abc = (a + b - c)(a2 + b2 + c2 - ab + bc + ca)
⇒ a3 + b3 - c3 + 3abc = (a + b - c){(a2 + b2 + c2 - (ab - bc - ca)}
दिलेले आहे:
a + b - c = 20
a2 + b2 + c2 = 152
(ab + bc - ca) शोधू:
⇒ (a + b - c)2 = a2 + b2 + c2 + 2ab - 2bc - 2ca
⇒ 202 = 152 + 2ab - 2bc - 2ca
⇒ 400 = 152 + 2(ab - bc - ca)
⇒ 248 = 2(ab - bc - ca)
⇒ ab - bc - ca = 124
पदावलीमध्ये ठेवू:
a3 + b3 - c3 + 3abc
⇒ (a + b - c){(a2 + b2 + c2 - (ab - bc - ca)}
= 20 × (152 - 124)
= 20 × 28
= 560
∴ a3 + b3 - c3 + 3abc चे मूल्य 560 आहे.
Last updated on Jul 8, 2025
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